MCQ 11 Mark
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
- A
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}+\lambda(\hat{\text{i}}+\hat{\text{j}})$
- ✓
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
- C
$\vec{\text{r}}=\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}+\lambda\hat{\text{k}}$
- D
$\vec{\text{r}}=\lambda\hat{\text{k}}$
AnswerCorrect option: B. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
Given,
a = (-1, 5, 4)
b = (0, 0, 1) [$\therefore$ 1 to plone z]
We know that,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
View full question & answer→MCQ 21 Mark
The cosines of the angle between any two diagonals of a cube is:
- ✓
$\frac{1}{3}$
- B
$\frac{1}{2}$
- C
$\frac{2}{3}$
- D
$\frac{1}{\sqrt{3}}$
AnswerCorrect option: A. $\frac{1}{3}$
$\frac{1}{3}$
View full question & answer→MCQ 31 Mark
The xy-plane divided the line joining the point (-1, 3, 4) and (2, -5, 6)
- A
Internally in the ratio 2 : 3
- ✓
Externally in the ratio 2 : 3
- C
Internally in the ratio 3 : 2
- D
Externally in the ratio 3 : 2
AnswerCorrect option: B. Externally in the ratio 2 : 3
Let the XY-plane divide the line segment joining points
P(-1, 3, 4) and Q(2, -5, 6) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(2)-1}{\text{k}+1},\frac{\text{k}(-5)+3}{\text{k}+1},\frac{\text{k}(6)+4}{\text{k}+1}\Big) $
On the XY-plane, the Z-coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(6)+4}{\text{k}+1}=0$
$\Rightarrow6\text{k}+4=0$
$\Rightarrow\text{k}=\frac{-2}{3}$
Thus, the XY-plane divides the line segment joining the given points in the ratio 2 : 3 externally.
View full question & answer→MCQ 41 Mark
A line OP where O = (0, 0, 0) makes equal angles with ox, oy, oz. The point on OP, which is at a distance of 6 units from O is:
- ✓
$\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
- B
$\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
- C
$-\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
- D
$-\big(6\sqrt{3},-6\sqrt{3},6\sqrt{3}\big)$
AnswerCorrect option: A. $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
$\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
View full question & answer→MCQ 51 Mark
Choose the correct answer from the given four options.
The locus represented by xy + yz = 0 is:
- A
A pair of perpendicular lines.
- B
A pair of parallel lines.
- C
A pair of parallel planes.
- ✓
A pair of perpendicular planes.
AnswerCorrect option: D. A pair of perpendicular planes.
We have, xy + yz = 0
⇒ xy = -yz
So, a pair of perpendicular planes.
View full question & answer→MCQ 61 Mark
The equation $x^2- x - 2 = 0$ in three dimensional space is represented by:
- ✓
A pair of parallel planes
- B
- C
A pair of perpendicular plane
- D
AnswerCorrect option: A. A pair of parallel planes
View full question & answer→MCQ 71 Mark
If $l, m, n$ are the direction cosines of a line, then:
- A
$l^2+ m^2+ 2n^2 = 1$
- B
$l^2+ 2m^2+ n^2 = 1$
- C
$2l^2+ m^2+ n^2 = 1$
- ✓
$l^2+ m^2+ n^2 = 1$
AnswerCorrect option: D. $l^2+ m^2+ n^2 = 1$
View full question & answer→MCQ 81 Mark
For every point P(x, y, z) on the x-axis (except the origin),
AnswerBoth Y and Z coordinates on each point of the x-axis are equal to zero.
The X-coordinate on the origin is also equal to zero.
Therefore, the Y and Z coordinates on each point of the x-axis, except the origin, are equal to zero,
While the X-coordinate is non-zero.
View full question & answer→MCQ 91 Mark
A straight line L on the xy-plane bisects the angle between OX and OY. What are the direction cosines of L:
- ✓
$\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
- B
$\Big(\frac{1}{2},\frac{\sqrt{3}}{2},0\Big)$
- C
$\big(0,0,1\big)$
- D
$\Big(\frac{2}{3},\frac{2}{3},\frac{1}{3}\Big)$
AnswerCorrect option: A. $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
L makes an angle $\frac{\pi}{4}$ with X and Y axis and $\frac{\pi}{2}$
$\therefore$ d.cs are $\Big(\cos\frac{\pi}{34},\cos\frac{\pi}{4},\cos\frac{\pi}{2}\Big)=\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
View full question & answer→MCQ 101 Mark
A line passes through the points (6, −7, −1) and (2, −3, 1). The direction cosines of the line so directed that the angle made by it with the positive direction of x-axis is acute, is?
- ✓
$\frac{2}{3},\frac{2}{3},-\frac{1}{3}$
- B
$-\frac{2}{3},\frac{2}{3},\frac{1}{3}$
- C
$\frac{2}{3}-\frac{2}{3},\frac{1}{3}$
- D
$\frac{2}{3},\frac{2}{3},\frac{1}{3}$
AnswerCorrect option: A. $\frac{2}{3},\frac{2}{3},-\frac{1}{3}$
Consider the problem
Let l, m, n are direction cosines of the given line.
then as it made an acute angle with x−axis,
Therefore, l > 0
The line passes through (6, −7, −1) and (2, −3, 1)
Therefore, its direction ratios are
6 − 2, −7 + 3, −1−1 or 2, −2, −1
Hence direction cosines of the line are given by $\frac{2}{3},\frac{2}{3},-\frac{1}{3}.$
View full question & answer→MCQ 111 Mark
The equation of the plane parallel to the lines x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z -4 and passing through the point (2, 3, 3) is:
AnswerLet a, b, c be the dirction ratios of the required plane.
The given line equation can be rewritten as
$\frac{\text{x}-1}{1}=\frac{\text{y}-\frac{5}{2}}{\frac{1}{2}}=\frac{\text{z}-0}{\frac{1}{2}}\ .....(1)$
$\frac{\text{x}-0}{\frac{1}{3}}=\frac{\text{y}-\frac{11}{4}}{\frac{1}{4}}=\frac{\text{z}-\frac{4}{3}}{\frac{1}{3}}\ .....(2)$
Since the required plane is parallel to the lines (1) and (2),
$\text{a}+\frac{\text{b}}{2}+\frac{\text{c}}{2}=0\Rightarrow2\text{a}+\text{b}+\text{c}=0....(3)$
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=0\Rightarrow4\text{a}+3\text{b}+4\text{c}=0....(4)$
Solving (3) and (4) using cross-multiplication method, we get
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=\lambda\text{(say)}$
$\Rightarrow\text{a}=\lambda,\text{b}=-4\lambda,\text{c}=2\lambda$
Now, the eqution of the plane whose direction ratios are $\lambda,-4\lambda,2\lambda$ and passing through the point.
$\lambda(\text{x}-2)+(-4\lambda)(\text{y}-3)+2\lambda(\text{z}-3)=0$
$\Rightarrow\text{x}-4\text{y}+2\text{z}+4=0$
View full question & answer→MCQ 121 Mark
If the directions cosines of a line are A, k, k, then:
AnswerCorrect option: D. $\text{k}=\frac{\sqrt{1}}{3}$ or $\frac{\sqrt{1}}{3}$
$\text{k}=\frac{\sqrt{1}}{3}$ or $\frac{\sqrt{1}}{3}$
View full question & answer→MCQ 131 Mark
Direction cosines of ray from P(1, −2, 4) to Q(−1, 1, −2) are:
AnswerCorrect option: D. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$
Given the points are P(1, −2, 4) and Q(−1, 1, −2).
Now the direction ratios of the ray PQ are (−1−1, 1 + 2, −2−4) = (−2, 3, −6).
The direction cosines of the line PQ will be
$\bigg(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\bigg)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big).$
View full question & answer→MCQ 141 Mark
The direction ratios of two lines AB, AC are 1, -1, -1 and 2, -1, 1. The direction ratios of the normal to the plane ABC are:
MCQ 151 Mark
The length of the $\perp$
- A
- B
$2\sqrt{3}$
- C
$\frac{2}{3}$
- ✓
View full question & answer→MCQ 161 Mark
The sine of the angle between the straight line $\frac{\text{x}-2}{3}=\frac{\text{y}-3}{4}=\frac{\text{z}-4}{5}$ and the plane 2x - 2y + z = 5 is:
AnswerCorrect option: C. $\frac{2\sqrt{3}}{5}$
$\frac{2\sqrt{3}}{5}$
View full question & answer→MCQ 171 Mark
The equation xy = 0 in three dimensional space is represented by:
- A
- ✓
Two plane are right angles
- C
A pair of parallel planes
- D
AnswerCorrect option: B. Two plane are right angles
Two plane are right angles
View full question & answer→MCQ 181 Mark
Choose the correct answer from the given four options.If the directions cosines of a line are $k, k, k,$ then:
AnswerCorrect option: D. $\text{k}=\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$
Since, direction cosines of a line are $k, k,$ and $k.$
$\therefore l = k, m = k$ and $n = k$
We know that$, l^2 + m^2 + n^2 = 1$
$\Rightarrow k^2 + k^2 + k^2 = 1$
$\text{k}^2=\frac{1}{3}$
$\therefore\text{k}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 191 Mark
If points (1, 2), (3, 5) and (0, b) are collinear the value of b is:
- ✓
$\frac{1}{2}$
- B
$\frac{7}{2}$
- C
$2$
- D
$-1$
AnswerCorrect option: A. $\frac{1}{2}$
Area $=\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-5)|$
As points are collinear, so area = 0
$\therefore\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-\text{5})|=0$
⇒ 5 − b + 3b − 6 = 0
⇒ = 1 = 2b
$\therefore\text{b}=\frac{1}{2}$
View full question & answer→MCQ 201 Mark
The straight line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
View full question & answer→MCQ 211 Mark
If a line makes 45°, 60° with positive direction of axes x and y then the angles it makes with the z-axis is:
MCQ 221 Mark
A line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x + y + z = 9$ at point $Q$. The length of the line segment $PQ$ equals:
- A
$1$
- B
$\sqrt{2}$
- ✓
$\sqrt{3}$
- D
$2$
AnswerCorrect option: C. $\sqrt{3}$
$\sqrt{3}$
View full question & answer→MCQ 231 Mark
If the direction ratios of two lines are given by 3lm - 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is:
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
AnswerCorrect option: D. $\frac{\pi}{2}$
$\frac{\pi}{2}$
View full question & answer→MCQ 241 Mark
If O is the origin, OP = 3 with direction ratios proportional to -1, 2, -2 then the coordinates of P are:
AnswerCorrect option: A. $(-1, 2,-2)$
Let the coordinates of P be (x, y, z). Then,
Direction ratios of OP = Coordinates of P-Coordinates of O-1, 2, 2 = (x - 0), (y - 0), (z - 0)
Thus, coordinates of P are (-1, 2, -2).
View full question & answer→MCQ 251 Mark
Direction ratio of line joining (2, 3, 4) and (-1, -2, 1), are:
AnswerThe direction ratio of the line joining A(2, 3, 4) and B(-1, -2, 1), are.
= (-1 - 2), (-2 - 3), (1 - 4)
= (-3, -5, -3)
View full question & answer→MCQ 261 Mark
The distance of the points (2, 1, -1) from the plane x - 2y + 4z - 9 is:
- A
$\frac{\sqrt{31}}{21}$
- B
$\frac{13}{21}$
- ✓
$\frac{13}{\sqrt{21}}$
- D
$\sqrt{\frac{\pi}{2}}$
AnswerCorrect option: C. $\frac{13}{\sqrt{21}}$
$\frac{13}{\sqrt{21}}$
View full question & answer→MCQ 271 Mark
A line makes the same angle $\theta$ with each of the $x$ and $z$ axis. If the angle $\beta$ which it makes with $y-$axis is such that $\sin^2\beta=3\sin^2\theta$ then $\cos^2\theta$ equals:
- ✓
$\frac{3}{5}$
- B
$\frac{1}{5}$
- C
$\frac{2}{3}$
- D
$\frac{2}{5}$
AnswerCorrect option: A. $\frac{3}{5}$
If a line makes the angle $\alpha,\beta,\gamma$ with $x, y, z$ axix respectively then
$l^2 + m^2 + n^2 = 1$
$\Rightarrow 2l^2 + m^2 = 1$ or $2n^2 + m^2 = 1$
$\Rightarrow2\cos^2\theta=1-\cos^2\beta (\alpha=\gamma=\theta)$
$2\cos^2\theta=\sin^2\beta$
$\Rightarrow2\cos^2\theta=3\sin^2\theta$
$\Rightarrow5\cos^2\theta=3$
View full question & answer→MCQ 281 Mark
The direction cosines of the line joining (1, -1, 1) and (-1, 1, 1) are:
AnswerCorrect option: C. $\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}}$
View full question & answer→MCQ 291 Mark
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
- A
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}+\lambda(\hat{\text{i}}+\hat{\text{j}})$
- ✓
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
- C
$\vec{\text{r}}=\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}+\lambda\hat{\text{k}}$
- D
$\vec{\text{r}}=\lambda\hat{\text{k}}$
AnswerCorrect option: B. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
Given,
a = (-1, 5, 4)
b = (0, 0, 1) [$\therefore$ 1 to plone z]
We know that,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
View full question & answer→MCQ 301 Mark
ox, oy are positive x-axis, positive y-axis respectively where O = (0, 0,0) The d.c.s of the llne which bisects $\angle\text{xoy}$ are:
- A
$1,1,0$
- ✓
$\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$
- C
$\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}$
- D
$0,0,1$
AnswerCorrect option: B. $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$
Equation of line bisecting XOY is x = y
$\therefore$ d.r.s are (1, 1, 0)
And thus d.c.s are $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
View full question & answer→MCQ 311 Mark
The direction cosines of the ray P(1, -2, 4) and Q(-1, 1, -2) are:
- A
$\big(-2, -3, -6\big)$
- B
$\big(2, -3, -6\big)$
- C
$\Big(\frac{2}{7},\frac{3}{7},\frac{6}{7}\Big)$
- ✓
$\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$
AnswerCorrect option: D. $\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$
P(1, -2, 4), Q(-1, 1, -2)
$\text{PQ}=\sqrt{(1-(1))^2 +(2-1)^2+(4-(-2))^{2}}$
$=\sqrt{4+9+36}$
$=\sqrt{49}=7\text{DC}$
$=\Big(\frac{-1-1}{7},\frac{1-(2)}{7},\frac{-2-4}{7}\Big)$
$=\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$
View full question & answer→MCQ 321 Mark
The direction ratios of the diagonal of the cube joining the origin to the opposite corner are $($when the $3$ concurrent edges of the cube are coordinate axes$)$.
AnswerCorrect option: B. $1, 1, 1$
Since, a cube is a symmetric figure, the vertex we are talking about will be at the diagonally opposite end of the origin.
i.e. it will be equally inclined to the three axes.
Let the side of the cube be a, then the corner opposite to origin will have coordinates $(a, a, a).$
Direction ratios of a line joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $(x_{2}− x_1, y_2− y_1, z_{2}− z_1)$
Then, direction ratios of two point $(0, 0, 0)$ and $(a, a, a)$ will be $(a − 0, a − 0, a − 0) = (a, a, a) = a(1, 1, 1)$
Hence, the direction ratios are $1, 1, 1.$
View full question & answer→MCQ 331 Mark
If a line has direction ratios 2, -1, -2, determine its direction cosines:
- A
$\frac{1}{3}, \frac{2}{3},\frac{-1}{3}$
- ✓
$\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
- C
$\frac{-2}{3}, \frac{1}{3}, \frac{2}{3}$
- D
AnswerCorrect option: B. $\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
Direction cosines are.
$=\frac{2}{2^2+(-1)^2+(-2)^2},\frac{1}{2^2+(-1)^2+(-2)^2},\frac{-2}{2^2+(-1)^2+(-2)^2}$
$=\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
View full question & answer→MCQ 341 Mark
The line x = 1, y = 2 is:
MCQ 351 Mark
The distance between the point (3, 4, 5) and the point where the line $\frac{\text{x}-3}{\text{1}}=\frac{\text{y}-4}{\text{2}}=\frac{\text{z}-5}{\text{2}}$ meets the plane x + y + z = 17 is:
AnswerThe coordinates of any point on the given line are of the from
$\frac{\text{x}-3}{\text{1}}=\frac{\text{y}-4}{\text{2}}=\frac{\text{z}-5}{\text{2}}=\lambda$
$\Rightarrow \text{x}=\lambda+3;\text{y}=2\lambda+4;\text{z}=2\lambda+5$
So, the coordinates of the point on the given line are $(\lambda+3,2\lambda+4,2\lambda+5)$
This point lies on the plane
x + y + z = 17
$\Rightarrow\lambda+3,2\lambda+4+2\lambda+5=17$
$\Rightarrow5\lambda=5$
$\Rightarrow\lambda=1$
So, the coordinates of the point are
$(\lambda+3,2\lambda+4,2\lambda+5)$
$=(1+3,2(1))+4,2(1)+5)$
$=(4,6,7)$
Now, the distance between the points (4, 6, 7) and (3, 4, 5) is
$\sqrt{(3+4)^2+(4-6)^2+(5-7)^2}$
$\sqrt{1+4+4}$
$=3\text{ units}$
View full question & answer→MCQ 361 Mark
If the projections of the line segment AB on the coordinate axes are 2, 3, 6, then the square of the sine of the angle made by AB with x = 0, is:
- A
$\frac{3}{7}$
- B
$\frac{3}{49}$
- C
$\frac{4}{7}$
- ✓
$\frac{40}{49}$
AnswerCorrect option: D. $\frac{40}{49}$
$\frac{40}{49}$
View full question & answer→MCQ 371 Mark
A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3) are the vertices of a tringle ABC. if the bisector of $\angle\text{ABC}$ meets BC at D, then coordinates of D are:
- ✓
$\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
- B
$\Big(-\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
- C
$\Big(\frac{19}{8},-\frac{57}{16},\frac{17}{16}\Big)$
- D
$\text{none of these}$
AnswerCorrect option: A. $\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
Since the bisector of $\angle\text{ABC}$ cannot meet BC, the solution of this quation is not possible.
Disclaimer: This quation is wrong, so the solution has not been provide.
View full question & answer→MCQ 381 Mark
The distance of the point $(-3, 4, 5)$ from the origin:
AnswerCorrect option: B. $5\sqrt{2}$
View full question & answer→MCQ 391 Mark
What is the sum of the squares ofdirection cosines of the line joining thepoints (1, 2, -3) and (-2, 3, 1):
- A
- ✓
- C
- D
$\frac{2}{\sqrt{26}}$
AnswerThe sum of the squares of direction cosines of the line is always 1
View full question & answer→MCQ 401 Mark
The distance of the line $\vec{\text{r}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$ from the plane $\vec{\text{r}}.(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}})=5$ is:
- A
$\frac{5}{3\sqrt{3}}$
- ✓
$\frac{10}{3\sqrt{3}}$
- C
$\frac{25}{3\sqrt{3}}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{10}{3\sqrt{3}}$
The given line passes through the point whose position vector is $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}.\vec{\text{n}}=\text{d}$ is given by
$\text{P}=\frac{\big|\vec{\text{a}}.\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}},\text{d}=5$
So, the required distance P is given by
$\text{P}=\frac{\Big|\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big),\big(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\big)-5\Big|}{\Big|\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\Big|}$
$=\frac{|2-10+3-5|}{\sqrt{1+25+1}}$
$=\frac{|-10|}{\sqrt{27}}$
$=\frac{10}{3\sqrt{3}}\text{units}$
View full question & answer→MCQ 411 Mark
The direction ratios of the normal to the plane $7x + 4y - 2z + 5 = 0$ are:
- ✓
$7, 4, -2$
- B
$7, 4, 5$
- C
$7, 4, 2$
- D
$4, -2, 5$
AnswerCorrect option: A. $7, 4, -2$
View full question & answer→MCQ 421 Mark
Ratio in which the xy-plane divided the join of (1, 2, 3) and (4, 2, 1) is:
AnswerSuppose the XY-plane divides the line segment joining the points P(1, 2, 3) and Q(4, 2, 1) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(4)+1}{\text{k}+1},\frac{\text{k}(2)+2}{\text{k}+1},\frac{\text{k}(1)+3}{\text{k}+1}\Big)$
The Z-coordinate of any point on the XY-plane is zero
$\Rightarrow\frac{\text{k}(1)+3}{\text{k}+1}=0$
$\Rightarrow\text{k}+3=0$
$\Rightarrow\text{k}=-3=\frac{-3}{1}$
Thus, the XY-plane divided the line segment joining the given points in the ratio 3 : 1 externally.
View full question & answer→MCQ 431 Mark
A normal to the plane x = 2 is:
AnswerThe plane x = 2 is perpendicular to x axis So the angle is $\frac{\pi}{2},\cos\frac{\pi}{2}=0$
0 The plane x = 2 is parallel to both y axis and z axis So the angle is (0, 1, 1)
View full question & answer→MCQ 441 Mark
Find the equation of the plane passing through the points $P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4):$
- A
$x + 2y = 0$
- B
$x - y - 2 = 0$
- C
$-x + 2y - 2 = 0$
- ✓
$x + y - 2 = 0$
AnswerCorrect option: D. $x + y - 2 = 0$
View full question & answer→MCQ 451 Mark
If $(0, 0),(a, 0)$ and $(0, b)$ are collinear, then:
- ✓
$ab = 0$
- B
$a = b$
- C
$a = −b$
- D
$a - b = c$
AnswerCorrect option: A. $ab = 0$
View full question & answer→MCQ 461 Mark
Which of the following triplets give the direction cosines of a line:
AnswerCorrect option: D. $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
If $l, m, n$ are the directions cosine of a line then $l^2+ m^2 + n^2 = 1$ Thus we get $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 471 Mark
The distance of the plane $2x - 3y + 6z + 7 = 0$ from the point $(2, -3, -1)$ is:
- A
$4$
- B
$3$
- ✓
$2$
- D
$\frac{1}{5}$
View full question & answer→MCQ 481 Mark
The equation of the plane through point $(1, 2, -3)$ which is parallel to the plane $3x - 5y + 2z = 11$ is given by:
- A
$3x - 5y + 2z - 13 = 0$
- B
$5x - 3y + 2z + 13 = 0$
- C
$3x - 2y + 5z + 13 = 0$
- ✓
$3x - 5y + 2z + 13 = 0$
AnswerCorrect option: D. $3x - 5y + 2z + 13 = 0$
View full question & answer→MCQ 491 Mark
If $P$ be the point $(2, 6, 3)$ then the equation of the plane trough $P,$ at right angles to $OP,$ where $′\ O\ ′$ is the origin is:
- A
$2x + 6y + 3z = 7$
- B
$2x − 6y + 3z = 7$
- C
$2x + 6y − 3z = 49$
- ✓
$2x + 6y + 3z = 49$
AnswerCorrect option: D. $2x + 6y + 3z = 49$
View full question & answer→MCQ 501 Mark
If $(0, 0),(a, 0)$ and $(0, b)$ are collinear, then:
- ✓
$ab = 0$
- B
$a = b$
- C
$a = −b$
- D
$a - b = c$
AnswerCorrect option: A. $ab = 0$
View full question & answer→MCQ 511 Mark
What are the direction cosines of a line which is equally inclined to the positive directions of the axes:
- ✓
$\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
- B
$\Big(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
- C
$\Big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\Big)$
- D
$\Big(\frac{1}{3},\frac{1}{3},\frac{1}{3}\Big)$
AnswerCorrect option: A. $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
We know sum of the squares of the direction cosines is one.
i.e. $\cos^2\alpha+\cos^2\gamma=1$
but its given that $\alpha=\beta=\gamma\therefore\cos^2\alpha=1$
$3\cos^2\alpha=1$
$\therefore\cos^2\alpha=\frac{1}{3}$
$\therefore$ Positive directions of the axes are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
View full question & answer→MCQ 521 Mark
The projections of a line segment on x, y and z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are:
- ✓
$13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
- B
$19;\frac{12}{19},\frac{4}{19},\frac{3}{19}$
- C
$11;\frac{12}{11},\frac{14}{11},\frac{3}{11}$
- D
$\text{None of these}$
AnswerCorrect option: A. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$
Let r be the length of the line segment. then,
$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$
$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$
$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$
$\Rightarrow\text{r}^2(1)=169$ [From (1)]
$\Rightarrow\text{r}=\sqrt{169}$
$\Rightarrow\text{r}=\pm13$
$\Rightarrow\text{r}=13$ (Since length cannot be negative)
(Since legth cannot be negative)
Substituting r = 13 in (2), we get
$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$ Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$
View full question & answer→MCQ 531 Mark
The image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0 is:
AnswerLet Q be the image of the point P(1, 3, 4) in the plane 2x - y +z + 3 = 0
Then PQ is normal to the plane.
So, the direction ratios of PQ are proportional to 2, -1, 1 equation of PQ is
Let the coordinates of Q be (2r + 1, -r + 3, r + 4)
Let R be the mid point of PQ.
Then,
$\text{R}=\Big(\frac{2\text{r}+1+1}{2},\frac{-\text{r}+3+3}{2},\frac{\text{r}+4+4}{2}\Big)$
$=\Big(\text{r}+1,\frac{-\text{r}+6}{2},\frac{\text{r}+8}{2}\Big)$
Since R lies in the plane 2x - y + z + 3 = 0,
$2(\text{r}+1)-\Big(\frac{-\text{r}+6}{2}\Big)+\frac{\text{r}+8}{2}+3=0$
⇒ 4r + 4 + r - 6 + r + 8 + 6 = 0
⇒ 6r + 12 = 0
⇒ r = -2
Substituting this in the coordinates of Q, we get
Q = (2r + 1, -r + 3, r + 4)
=(2 (-2) + 1, 2 + 3, -2 + 4)
=(-3, 5, 2).
View full question & answer→MCQ 541 Mark
The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:
AnswerWe have
$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$
$\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$
The direction of the given lines are propotional to 2, 5, 4 are 1, 2, -3.
The given lines are parallel to the vectors $\vec{\text{b}}_1=2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}.$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$
$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$
$\Rightarrow\theta=90^\circ$
View full question & answer→MCQ 551 Mark
If P(x, y, z) moves such that x = 0, z = 0 then the locus of P is the line whose d.cs are:
AnswerWhen P moves then x = 0, z = 0 but y is not given. Let y = y Then the coordinates of the point will be (0, y, 0) Now, direction cosines with respect to (0, y, 0) is given by.
$\cos\alpha=\frac{0}{0^2+\text{y}^2+0^2}=\frac{0}{\text{y}}=0$
$\cos\beta=\frac{\text{y}}{0^2+\text{y}^2+0^2}=\frac{\text{y}}{\text{y}}=1$
$\cos\gamma=\frac{{0}}{0^2+\text{y}^2+0^2}=\frac{{0}}{\text{y}}=0$
The direction cosines are 0, 1, 0
View full question & answer→MCQ 561 Mark
A rectangular parallelopiped is formed by planes drawn through the point (5, 7, 9) and (2, 3, 7) parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is:
AnswerThe give point (5, 7, 9) and (2, 3, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates.
Edges of the paralleloppiped = |5 - 2|, |7 - 3|, |9 - 7|
=3, 4, 2.
View full question & answer→MCQ 571 Mark
If $\alpha,\beta$ and $\gamma$ are the angles which a half ray makes with the positive direction of the axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:
AnswerGiven expression, $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$
$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$
$(\because\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$
View full question & answer→MCQ 581 Mark
The direction ratios of the line joining the points $A(4, −3, 7)$ and $B(1, 3, 5)$ are:
AnswerCorrect option: B. $(3, −6, 2)$
View full question & answer→MCQ 591 Mark
The eqution of the plane $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$ in scalar product from is:
- ✓
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
- B
$\vec{\text{r}}.(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=7$
- C
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})=7$
- D
$\text{None of these}$
AnswerCorrect option: A. $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represents a plane passing through a point whose position vectors is $\vec{\text{a}}$ and parrallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}};\ \vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\1&-2&3\end{vmatrix}$
$=5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
View full question & answer→MCQ 601 Mark
Direction cosines of ray from P(1, -2, 4) to Q(-1, 1, -2) are:
AnswerCorrect option: D. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$
Given the points are P(1, -2, 4) and Q(-1, 1, -2) Now the direction
ratios of the ray PQ are (-1 - 1, 1 + 2, -2 - 4) = (-2, 3, -6)
The direction cosines of the line PQ will be
$\Big(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\Big)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big)$
View full question & answer→MCQ 611 Mark
The direction ratios of the normal to the plane $7x + 4y - 2z + 5 = 0$ are:
- ✓
$7, 4, -2$
- B
$7, 4, 5$
- C
$7, 4, 2$
- D
$4, -2, 5$
AnswerCorrect option: A. $7, 4, -2$
View full question & answer→MCQ 621 Mark
The angle between the two diagonals of a cube is:
AnswerCorrect option: A. $30^\circ$
:
Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, Consider the diagonals OP and AR.
Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, e.i. a, a, a and -a, a, a, respectivelly.
Let $\theta$ be the angle between OP and AR. Then,
$\cos\theta=\frac{\text{a}\times-\text{a}+\text{a}\times\text{a}+\text{a}\times\text{a}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{(-\text{a})^2+\text{a}^2+\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{-\text{a}+\text{a}^2+\text{a}^2}{\sqrt{3\text{a}^2}\sqrt{3\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{1}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$
Similarly, the angles between other pairs of the diagonals are equal to $\cos^{-1}\Big(\frac{1}{3}\Big)$ as the angle between any two diagonals. View full question & answer→MCQ 631 Mark
The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:
- ✓
$3,1,-2$
- B
$2,-4,1$
- C
$\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$
- D
$\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$
AnswerCorrect option: A. $3,1,-2$
We have
x - y + z - 5 = 0 = x - 3y - 6
⇒ x - 3y - 6=0
x - y + z - 5 = 0
⇒ x = 3y + 6 .....(1)
x - y + z - 5 = 0.....(2)
From (1) and (2), we get
3y + 6 - y + z - 5 = 0
⇒ 2y + z + 1 = 0
$\Rightarrow\text{y}=\frac{-\text{z}-1}{2}$
$\text{y}=\frac{\text{x}-6}{3}$ [From (1)]
$\therefore\frac{\text{x}-6}{3}=\text{y}=\frac{-\text{z}-1}{2}$
So, the given equation can be re-witten as
$\frac{\text{x}-6}{3}=\frac{\text{y}}{1}=\frac{\text{z}+1}{-2}$
Hence, the direction ratios the given line are proportional to 3, 1, -2.
View full question & answer→MCQ 641 Mark
The plane $2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$ passes through the intersection of the planes:
- ✓
- B
- C
2x - y + 3z = 0 and y - 3z = 0
- D
AnswerThe given plane is
$2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$
$\Rightarrow(2\text{x}-\text{y})+\lambda(-\text{y}+3\text{z})=0$
So, this plane passes through the intersection of the planes
2x - y = 0 and -y + 3z = 0
⇒ 2x - y = 0 and y - 3z = 0.
View full question & answer→MCQ 651 Mark
Which of the following represents direction cosines of the line:
- A
$0,\frac{1}{\sqrt{2}},\frac{1}{2}$
- B
$0,\frac{-\sqrt{3}}{2},\frac{1}{\sqrt{2}}$
- ✓
$0,\frac{\sqrt{3}}{2},\frac{1}{2}$
- D
$\frac{1}{2},\frac{1}{2},\frac{1}{2}$
AnswerCorrect option: C. $0,\frac{\sqrt{3}}{2},\frac{1}{2}$
If direction cosine of a line is $l, m, n$ then
$l^2 + m^2 + n^2 = 1$
$=0^2+\Big(\frac{\sqrt{3}}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2=1$
The correct answer from the given alternative is $(c) \ 0,\frac{\sqrt{3}}{2},\frac{1}{2}$
View full question & answer→MCQ 661 Mark
If $\alpha,\beta,\gamma$ are the angles which a directed line makes with the positive directions of the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:
AnswerThe direction cosines of the line are
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Now, $\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow1-\sin^2\alpha+1-\sin^2\beta+1-\sin^2\gamma=1$
$\Rightarrow\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$
View full question & answer→MCQ 671 Mark
The reflection of the point $(\text{a}, \beta, \gamma) $ in the $xy-$plane is:
AnswerCorrect option: D. $(\alpha,\beta,\gamma)$
View full question & answer→MCQ 681 Mark
What are the direction ratios of the line if it passes through the intersection of the planes x = 3z + 4 and y = 2z - 3:
AnswerEquations of the planes are x = 3z + 4 and y = 2z - 3
$\therefore$ The equation of the plane passing through the line of intersection of these planes is x = 3z + 4 and y = 2z - 3
Thus The direction Ratios of the equation passes through intersection of the planes is (3, 2, 1).
View full question & answer→MCQ 691 Mark
The product of the d.cs of the line which makesequal angles with ox, oy, oz is:
- A
$1$
- B
$\sqrt{3}$
- ✓
$\frac{1}{3\sqrt{3}}$
- D
$\frac{1}{\sqrt{3}}$
AnswerCorrect option: C. $\frac{1}{3\sqrt{3}}$
$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$
$\Rightarrow3\cos^2(\alpha)=1$
$\Rightarrow\cos\alpha=\underline{+}\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 701 Mark
A line passes through the points $(6, -7, -1)$ and $(2, -3, 1)$. What are the direction ratios of the line?
- A
$(4, −4, 2)$
- B
$(4, 4, 2)$
- ✓
$(−4, 4, 2)$
- D
$(2, 1, 1)$
AnswerCorrect option: C. $(−4, 4, 2)$
Direction ratios of a line passing through points $(x_1, y_1, z_1)$ and $(x_{2}, y_{2}, z_{2})$ are represented by $±(x_1−x_{2}, y_{1}−y_{2}, z_{1}−z_{2})$
Hence for the given line, direction ratios are $(6 − 2, −7−(−3), −1−1)$
$\Rightarrow ±(4, −4, −2)$
$\Rightarrow (−4, 4, 2)$ or $(4, −4, −2)$
View full question & answer→MCQ 711 Mark
If a line makes angle $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with x-axis and y-axis respectively, then the angle made by the line with z-axis is:
- A
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{3}$
- C
$\frac{\pi}{4}$
- D
$\frac{5\pi}{12}$
AnswerCorrect option: B. $\frac{\pi}{3}$
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axcs, then $\cos2\alpha+\cos2\beta+\cos2\gamma=1.$
Here,
$\alpha=\frac{\pi}{3}$
$\beta=\frac{\pi}{4}$
Now,
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\gamma=1$
$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\gamma=1 $
$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$
$\Rightarrow\cos^2\gamma=\frac{1}{4}$
$\Rightarrow\cos\gamma=\frac{1}{2}$
$\Rightarrow\gamma=\frac{\pi}{3}$
View full question & answer→MCQ 721 Mark
The direction cosines of the straight linegiven by the planes x = 0 and z = 0 are:
AnswerGiven, x = z = 0
It represents Z-axis
$\therefore$ Direction cosines = (0, 1, 0)
View full question & answer→MCQ 731 Mark
Choose the correct answer from the given four options.
The sine of the angle between the straight line $\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$ and the plane $2\text{x}-2\text{y}+\text{z}=5$ is:
- A
$\frac{10}{6\sqrt{5}}$
- B
$\frac{4}{5\sqrt{2}}$
- C
$\frac{2\sqrt{3}}{5}$
- ✓
$\frac{\sqrt{2}}{10}$
AnswerCorrect option: D. $\frac{\sqrt{2}}{10}$
We have, the equation of line as
$\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$
This line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
Equation of plane is $2\text{x}-2\text{y}+\text{z}=5$
Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Its angle between line and plane is $'\theta'.$
Then $\sin\theta=\frac{|\vec{\text{b}}\cdot{\vec{\text{b}}}|}{|{\vec{\text{b}}}||{\vec{\text{b}}}|}$
$=\frac{\big|(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}})\cdot(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})\big|}{\sqrt{3^2+4^2+5^2}\sqrt{4+4+1}}$
$=\frac{|6-8+5|}{\sqrt{50}\sqrt{9}}$
$=\frac{3}{15\sqrt{2}}=\frac{1}{5\sqrt{2}}$
$\sin\theta=\frac{\sqrt{2}}{10}$
View full question & answer→MCQ 741 Mark
The three points $\text{ABC}$ have position vectors $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$ are collinear then $(x, y):$
- ✓
$(2, -3)$
- B
$(-2, 3)$
- C
$(-2, -3)$
- D
$(2, 3)$
AnswerCorrect option: A. $(2, -3)$
View full question & answer→MCQ 751 Mark
The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:
- A
$\sqrt{30}$
- B
$2\sqrt{30}$
- C
$5\sqrt{30}$
- ✓
$3\sqrt{30}$
AnswerCorrect option: D. $3\sqrt{30}$
We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$
$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$ where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Also, line (2) passes through the point (3, -7, 6) and has direction ratios proprtional to -3, 2, 4.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2&4\end{vmatrix}$
$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$
$=\sqrt{36+225+9}$
$=\sqrt{270}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$
$=36+225+9$
$=270$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{270}{\sqrt{270}}\Big|$
$=\sqrt{270}$
$=3\sqrt{30}$
View full question & answer→MCQ 761 Mark
The projection of a directed line segment on the co-ordinate axes are 12, 4, 3, then the direction cosines of the line are:
- A
$\frac{-12}{13},\frac{-4}{13},\frac{-3}{13}$
- ✓
$\frac{12}{13},\frac{4}{13},\frac{3}{13}$
- C
$\frac{12}{13},\frac{-4}{13},\frac{3}{13}$
- D
$\frac{12}{13},\frac{4}{13},\frac{-3}{13}$
AnswerCorrect option: B. $\frac{12}{13},\frac{4}{13},\frac{3}{13}$
x = 12, y = 4, z = 3
Direction cosines =
$\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{y}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2}$
$=\frac{12}{13},\frac{4}{13},\frac{3}{13}$
View full question & answer→MCQ 771 Mark
The length of perpendicular from the origin to the plane which makes intercepts $\frac{1}{3},\frac{1}{4},\frac{1}{5}$ respectively on the coordinate axes is:
- ✓
$\frac{1}{\sqrt[5]{2}}$
- B
$\frac{1}{10}$
- C
$\sqrt[5]{2}$
- D
$5$
AnswerCorrect option: A. $\frac{1}{\sqrt[5]{2}}$
Equation of plane $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$3x + 4y + 5z − 1 = 0
diatance from origin $\frac{1}{\sqrt{150}}=\frac{1}{\sqrt[5]{2}}$
View full question & answer→MCQ 781 Mark
For every point P(x, y, z) on the xy-plane,
AnswerThe Z-coordinate of every point on the XY-plane is zero.
View full question & answer→MCQ 791 Mark
The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
AnswerWe have
$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$
Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular to the given line.
Now,
$3\text{x}+4\text{y}+0\text{z}=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).
Hence, the given line is perpendicular to z-axis.
View full question & answer→MCQ 801 Mark
The d.rs of the lines x = ay + b, z = cy + d are:
AnswerGiven x = ay + b and z = cy + d
$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\text{y}$ and $\frac{\text{z}-\text{d}}{\text{c}}=\text{y}$
$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{c}$
Therefore Drs of given line is a, 1, c
View full question & answer→MCQ 811 Mark
Choose the correct answer.Distance between the two planes$:\ 2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is:
AnswerCorrect option: D. $\frac{2}{\sqrt{29}}\ \text{units}.$
Equation of one plane is $2x + 3y + 4z = 4$
$\Rightarrow 2x + 3y + 4z - 4 = 0$
Equation of second plane is $4x + 6y + 8z = 12$
$\Rightarrow 4x + 6y + 8z - 12 = 0$
$\Rightarrow 2x + 3y + 4z - 6 = 0$
Here $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2},\ \frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2},\ \frac{\text{c}_1}{\text{c}_2}=\frac{4}{8}=\frac{1}{2}$
Since, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
therefore, the given two lines are parallel.
We know that the distance of parallel lines $=\frac{|\text{d}_1-\text{d}_2|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{|-4-(-6)|}{\sqrt{(2)^2+(3)^2+(4)^2}}$
$=\frac{|-4+6|}{\sqrt{4+9+16}}$
$=\frac{2}{\sqrt{29}}$
Therefore, option $(D)$ is correct.
View full question & answer→MCQ 821 Mark
The following lines are $\hat{\text{r}}=\Big(\hat{\text{i}}+\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\Big)+\mu\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
- ✓
- B
Skew$-$lines
- C
$Co-$planar lines
- D
View full question & answer→MCQ 831 Mark
Are the points $(1, 1), (2, 3)$ and $(8, 11)$ collinear?
AnswerArea of triangle formed by these vertices is,
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$
Applying $R_2 \rightarrow R_2 − R_1, R_{3}\rightarrow R_{3 }− R1$
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$
Hence points are non collinear
View full question & answer→MCQ 841 Mark
The direction cosines of a line which is equally inclined to axes, is given by:
AnswerCorrect option: B. $\underline{+}\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 851 Mark
Let a vector $\vec{\text{r}}$ make angles $60^\circ , 30^\circ $ with it and $y-$axes respectively. Find the angle $\vec{\text{r}}$ make with $z-$axis:
- A
$30^\circ $
- B
$60^\circ $
- ✓
$90^\circ $
- D
$120^\circ $
AnswerCorrect option: C. $90^\circ $
View full question & answer→MCQ 861 Mark
If the x-coordinate of a point P on the join of Q(2, 2, 1) and R(5, 1, -2) is 4, then its z-coordinate is:
AnswerSuppose the point P divided the line segment joining the point Q(2, 2, 1) and R(5, 1, -2) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(5)+2}{\text{k}+2},\frac{\text{k}(1)+2}{\text{k}+1},\frac{\text{k}(-2)+1}{\text{k}+1}\Big) $
On the XY-plane, the Z-coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(5)+2}{\text{k}+2}=4$
$\Rightarrow5\text{k}+2=4(\text{k}+1)$
$\Rightarrow\text{k}=2$
Now,
Z-coordinate of P $=\frac{\text{k}(-2)+1}{\text{k}+1}$
$\frac{2(-2)+1}{2+1}\ [\text{Substituting} \text{ k}=2]$
$=-1$
View full question & answer→MCQ 871 Mark
If P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divided PQ in the ratio:
AnswerSuppose the point R divides PQ in the ratio $\lambda:1$.
Coordinates of R are $\Big(\frac{5\lambda+3}{\lambda+1},\frac{4\lambda+2}{\lambda+1},\frac{-6\lambda-4}{\lambda+1}\Big)$.
But the coordinates of R are (9, 8, -10).
$\therefore\frac{5\lambda+3}{\lambda+1}=9,\frac{4\lambda+2}{\lambda+1}=8$ and $\frac{-6\lambda-4}{\lambda+1}=-10$
From each of these equations, we get
$\lambda=-\frac{3}{2}$
$\therefore$ R divided PQ in the ratio 3 : 2 externally.
View full question & answer→MCQ 881 Mark
How many lines through the origin in make equal angles with the coordinate axis:
MCQ 891 Mark
The locus of $xy + yz = 0$ is:
- A
- B
- C
A pair of parallel planes
- ✓
A pair of perpendicular planes
AnswerCorrect option: D. A pair of perpendicular planes
View full question & answer→MCQ 901 Mark
What are the direction ratios of normal to the plane 2x - y + 2z + 1 = 0:
AnswerDirection ratios of normal to the plane ax + by + cz + d = 0, are
a, b, c. So, here in the question the given plane is 2x - y + 2z + 1
= Thus, the direction ratios are 2, -1, 2
View full question & answer→MCQ 911 Mark
Choose the correct answer from the given four options. The reflection of the point $(\alpha,\beta,\gamma)$ in the xy–plane is:
AnswerCorrect option: D. $(\alpha,\beta,-\gamma)$
In XY-plane, only the sign of z coordinate of the point got changed after the reflection. Therefore, the reflection of the point $(\alpha,\beta,\gamma)$ is $(\alpha,\beta,-\gamma).$
View full question & answer→MCQ 921 Mark
The points $(k − 1, k + 2), (k, k + 1), (k + 1, k)$ are collinear for:
AnswerCorrect option: A. any value of $k$
View full question & answer→MCQ 931 Mark
If the points $(p, 0), (0, q)$ and $(1, 1)$ are collinear, then $\frac{1}{\text{p}}+\frac{1}{\text{q}}$ is equal to:
AnswerAs the points are collinear, the slope of the line joining
any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points $(x_1, y_1)$ and $\text{x}_2,\text{y}_2=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}$
So, slope of the line joining $(p, 0), (0, q) =$ Slope of the line joining
$(0, q)$ and $(1, 1)$
$\frac{\text{q}-0}{0-\text{p}}=\frac{1-\text{q}}{1-\text{p}}$
$-\frac{\text{q}}{\text{p}}=1-\text{q}$
Dividing both sides by $q,$
$-\frac{1}{\text{p}}=\frac{1}{\text{q}}-1$
$\Rightarrow\frac{1}{\text{p}}+\frac{1}{\text{q}}=1$
View full question & answer→MCQ 941 Mark
$(2, -3, -1) 2x - 3y + 6z + 7 = 0:$
- A
$4$
- B
$3$
- ✓
$2$
- D
$\frac{1}{5}$
View full question & answer→MCQ 951 Mark
The plane $x + y = 0:$
AnswerCorrect option: C. Passes through $z-$axis
View full question & answer→MCQ 961 Mark
The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:
AnswerThe equation of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$=\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
Thus, the two lines are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and pass through the points (0, 0, 0) and (1, 2, 3).
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$
View full question & answer→MCQ 971 Mark
Choose the correct answer The planes : $2x – y + 4z = 5$ and $5x – 2.5y + 10z = 6$ are :
AnswerEquation of the given planes are $2x - y + 4z = 5(a_1x + b_1y + c_1z + d = 0)$
and $5x - 2.5y + 10z = 6(a_2x + b_2y + c_2z + d = 0)$
For perpendicular $a_1a_2 + b_1b_2 + c_1c_2 = 2(5) + (-1)(-2.5) + 4(10) $
$= 10 + 2.5 + 40 = 52.5$
$\because\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2\neq0$
$\therefore$ Planes are not perpendicular.
For parallel $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{5},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-1}{-2.5}$
$=\frac{10}{25}=\frac{2}{5},\ \frac{\text{c}_1}{\text{c}_2}=\frac{4}{10}=\frac{2}{5}$
$\because\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ given planes are parallel.
Therefore, option $(B)$ is correct.
Parallel.
View full question & answer→MCQ 981 Mark
If a line makes angles $\alpha,\beta,\gamma$ with the axis then $\cos 2\alpha+ \cos 2\beta +\cos 2\gamma=$
View full question & answer→MCQ 991 Mark
If the lines $\text{ x - }\frac{2}{1} =\text{y}-\frac{2}{1} =\text{z}-\frac{4}{\text{k}} $ and $\text{x}-\frac{1}{\text{k}} = \text{y}-\frac{4}{2} = \text{z}-\frac{5}{1} $ are coplanar, then $k$ can have:
View full question & answer→MCQ 1001 Mark
Choose the correct answer from the given four options.
The distance of the plane $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1$ from the origin is:
AnswerThe general equation of a plane in vector form is given by $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$
Where d is the distance of the plane from the origin.
Comparing $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ and $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1,$ we get
Therefore, the distance of the plane $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1$ from the origin is 1.
View full question & answer→MCQ 1011 Mark
The direction cosines $l, m, n$ of two lines are connected by the relations $l + m + n = 0, lm = 0,$ then the angle between them is:
- ✓
$\frac{\pi}{3}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{2}$
- D
$0$
AnswerCorrect option: A. $\frac{\pi}{3}$
View full question & answer→MCQ 1021 Mark
Find the value of $p$ for which the points $(−5, 1), (1, p)$ and $(4, −2)$ are collinear.
AnswerThe given points are $A(−5, 1), B(1, p)$ and $C(4, −2)$
We have $(x_{1}= −5, y_{1 }= 1),(x_2 = 1, y_2 = p)$ and $(x_3 = 4, y_{3} = −2)$
The given points $A, B$ and $C$ are collinear
Therefore, $x_1(y_2 − y_3) + x_2(y_{3}− y_{1}) + x_3(y_1 − y_2) = 0$
$\Rightarrow (−5)⋅(p + 2) + 1⋅(−2−1) + 4⋅(1 − p) = 0$
$\Rightarrow (−5p − 10 − 3 + 4 − 4p) = 0$
$\Rightarrow −9p = −9$
$\Rightarrow p = −1$
Hence, $p = −1$
View full question & answer→MCQ 1031 Mark
If a line has the direction ratio 18, 12, 4, then its direction cosines are:
- ✓
$\frac{9}{11},\frac{6}{11},\frac{2}{11}$
- B
$\frac{9}{13},\frac{6}{13},\frac{2}{13}$
- C
$\frac{9}{7},\frac{6}{7},\frac{2}{7}$
- D
AnswerCorrect option: A. $\frac{9}{11},\frac{6}{11},\frac{2}{11}$
Dr's of the line are : 18, 12, 4
$\text{Dc's}=\frac{18}{\sqrt{18^2+12^2+4^2}},\frac{12}{\sqrt{18^2+12^2+4^2}},\frac{4}{\sqrt{18^2+12^2+4^2}}$
$=\frac{18}{22},\frac{12}{22},\frac{4}{22}$
$=\frac{9}{11},\frac{6}{11},\frac{2}{11}$
View full question & answer→MCQ 1041 Mark
The direction cosines of any normal to the $xy$ plane are:
- A
$1, 0 ,0$
- B
$0, 1, 0$
- C
$1, 1, 0$
- ✓
$1, 1, 0$
AnswerCorrect option: D. $1, 1, 0$
View full question & answer→MCQ 1051 Mark
A plane passing through (−1, 2, 3) and whose normal makes equal angle with the coordinate axes is:
AnswerSince normal makes equal angles with coordinate axis.
So, it intercept with all the axis will be same. So equation of plane will be
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}+\frac{\text{x}}{\text{a}}=1$
⇒ x + y + z = a
Now, it passes through (−1, 2, 3), so
−1 + 2 + 3 = a
⇒ a = 4
⇒ x + y + z − 4 = 0
View full question & answer→MCQ 1061 Mark
If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of a vector $\vec{\text{a}}$ then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
View full question & answer→MCQ 1071 Mark
If the three points A(1, 6), B(3, −4) and C(x, y) are collinear, then the equation satisfying by x and y is:
AnswerSince, the points A(1, 6), B(3, −4) and C(x, y) are colinear
$\therefore\begin{vmatrix}1&6&1\\3&-4&1\\\text{x}&\text{y}&1\end{vmatrix}=0$
⇒ 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0
⇒ 10x + 2y − 22 = 0
⇒ 5x + y − 11 = 0
View full question & answer→MCQ 1081 Mark
Choose the correct answer from the given four options.
Distance of the point $(\alpha,\beta,\gamma)$ from y-axis is:
AnswerCorrect option: D. $\sqrt{\text{a}^2+\gamma^2}$
Required distance $=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}$
$=\sqrt{\alpha+\gamma^2}$
View full question & answer→MCQ 1091 Mark
Distance of the point $(\alpha, \beta, \gamma)$ from $y-$axis is:
AnswerCorrect option: D. $\sqrt{\alpha^2+\gamma^2}$
View full question & answer→MCQ 1101 Mark
The direction cosines of the line passing through $P(2, 3, -1)$ and the origin are:
- A
$\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
- B
$\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
- ✓
$\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
- D
$\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{-1}{\sqrt{14}}$
AnswerCorrect option: C. $\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
View full question & answer→MCQ 1111 Mark
A line makes angles $\alpha,\beta,\gamma$ with the positive direction of the axes of reference. The value of $\cos2\alpha+\cos2\beta+\cos2\gamma$ is:
Answer$\cos^2\alpha+\cos^2\beta+\cos^2\text{r}=1\cos2\alpha+\cos^2\beta+\cos2\text{r}$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\text{r}-1$
$=2(\cos^2\alpha+\cos^2\beta\cos{\text{r}})-3=2(1)-3=-1$
View full question & answer→MCQ 1121 Mark
A plane meets the coordinate axes at A, B, and C such that the centroid of $\triangle{\text{ABC}}$ is the point (a, b, c) if
the eqution of the plane is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=\text{k}$,then k =
AnswerLet and be the interceots of the given plane on the coordinate axes.
Then, the plane meets the coordinate axes at
$\text{A}(\alpha,0,0),\text{B}(0,\beta,0)$ and $\text{C}=(0,0,\gamma)$
Given that the centroid of the triangle = (a, b, c)
$\Rightarrow\Big(\frac{\alpha+0+0}{3},\frac{0+\beta+0}{3},\frac{0+0+\gamma}{3}\Big)=(\text{a}+\text{b}+\text{c})$
$\Rightarrow\Big(\frac{\alpha}{3},\frac{\beta}{3},\frac{\gamma}{3}\Big)=(\text{a},\text{b},\text{c})$
$\Rightarrow\frac{\alpha}{3}=\text{a},\frac{\beta}{3}=\text{b},\frac{\gamma}{3}=\text{c}$
$\Rightarrow\alpha=3\text{a},\beta=3\text{b},\gamma=3\text{c}\ .....(1)$
Equation of the plane whose intercepts on the coordinate axes are $\alpha,\beta$ and $\gamma$ is
$\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1\ [\text{From (1)}]$
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=3$
View full question & answer→MCQ 1131 Mark
The direction ratios of the line 6x - 2 = 3y + 1 = 2z - 2 are:
- A
$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
- B
$\frac{1}{\sqrt{14}},\frac{12}{\sqrt{14}},\frac{3}{\sqrt{14}}$
- ✓
$1, 2, 3$
- D
AnswerCorrect option: C. $1, 2, 3$
6x - 2 = 3y + 1 = 2x - 2
$6\big(\text{x}-\frac{2}{6}\big)=3(\text{y}+\frac{1}{3}\big)=2(\text{x}-\frac{2}{2}\big)$
$\frac{\big(\text{x}-\frac{1}{3}\big)}{1}=\frac{\big(\text{y}+\frac{1}{3}\big)}{2}=\frac{(\text{x}-1)}{3}$
Line will be passing through the poits, $\Big(\frac{1}{3},-\frac{1}{3},1\Big)$
and parallel to the line having direction ratios is 1, 2, 3
View full question & answer→MCQ 1141 Mark
If the projections of the line segment $AB$ on the coordinate axes are $12, 3, k$ and $AB = 13$ then $k^2 - 2k + 3$ is equal to:
AnswerLet $a, b, c$ be the projection of a line on the coordinate axes.
Then the length of the line given by $\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
Here we have $12^2 + 3^2 + k^2 = 169$
$\Rightarrow\text{k}=\underline{+}4$
Thus $k^2 - 2k + 3 = 11$ or $27$.
View full question & answer→MCQ 1151 Mark
What are the DR's of vector parallel to (2, −1, 1) and (3, 4, −1)?
AnswerRequired DR's are (3 − 2, 4 + 1, −1−1) ie, (1, 5, −2)
View full question & answer→MCQ 1161 Mark
Choose the correct answer from the given four options.
The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2), is equal to:
AnswerWe have, A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2)
$\therefore\overrightarrow{\text{AB}}=(2-0)\hat{\text{i}}+(3-4)\hat{\text{j}}+(-1-1)\hat{\text{k}}$
$=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{BC}}=(4-2)\hat{\text{i}}+(5-3)\hat{\text{j}}+(0-0)\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{CD}}=(2-4)\hat{\text{i}}+(6-5)\hat{\text{j}}+(2-0)\hat{\text{k}}$
$=-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{DA}}=(0-2)\hat{\text{i}}+(4-6)\hat{\text{j}}+(1-2)\hat{\text{k}}$
Thus quadrilateral formed is parallelogram.
Area of quadrilateral ABCD
$=\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{BC}}\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-1&-2\\2&2&1 \end{vmatrix}$
$=|3\vec{\text{i}}-6\vec{\text{j}}+6\vec{\text{k}}|$
$=\sqrt{9+36+36}$
$=9\text{ sq.units}$
View full question & answer→MCQ 1171 Mark
Area of $\triangle\text{ABC}$ is:
AnswerLine PA: $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{1}$
Line PB: $\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-6}{1}$
Line PC: $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{-2}$
Then $\text{A}\Big(\frac{7}{2},-\frac{1}{2},\frac{17}{2}\Big)$
$\text{B}\Big(\frac{17}{2},-13,-\frac{3}{2}\Big)$
$\text{C}\Big(-14,\frac{19}{2},21\Big)$
Hence area of $\triangle\text{ABC}=\frac{225\sqrt{14}}{8},$ volume of tetrahedron
$\text{PABC}=\frac{125}{8}\text{cubic units}$
View full question & answer→MCQ 1181 Mark
The points with position vectors 60i + 3j, 40i − 8j and ai − 52j are collinear if:
AnswerDenoting a,b,c by the given vectors respectively
These vectors will be collinear if there is some constant k such that c − a = K(b − a)
⇒ a − 60 = −20K and −55 = −11K
⇒ a = −100 + 60 = −40
View full question & answer→MCQ 1191 Mark
The angle made by line $\text{r}[\cos\theta−3\sin\theta]=5 $ with initial line is:
AnswerGiven equation
$\text{r}[\cos\theta−3\sin\theta]=5 $
$\text{x}−\sqrt{3}\text{y}=5$
Slope of the line is $\tan\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=30^\circ$
Hence, the line makes an angle of 30° with the initial line.
View full question & answer→MCQ 1201 Mark
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
- A
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}+\lambda(\hat{\text{i}}+\hat{\text{j}})$
- ✓
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
- C
$\vec{\text{r}}=\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}+\lambda\hat{\text{k}}$
- D
$\vec{\text{r}}=\lambda\hat{\text{k}}$
AnswerCorrect option: B. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
Given,
a = (-1, 5, 4)
b = (0, 0, 1) [$\therefore$ 1 to plone z]
We know that,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
View full question & answer→MCQ 1211 Mark
The equation of the plane passing through $(2, −3, 1)$ and is normal to the line joining the points $(3, 4, −1)$ and $(2, −1, 5)$ is given by:
- ✓
$x + 5y − 6z + 19 = 0$
- B
$x − 5y + 6z − 19 = 0$
- C
$x + 5y + 6z + 19 = 0$
- D
$x − 5y − 6z − 19 = 0$
AnswerCorrect option: A. $x + 5y − 6z + 19 = 0$
View full question & answer→MCQ 1221 Mark
The sum of the squares of sine of the angles made by the line $AB$ with $\text{OX, OY, OZ}$ where $O$ is the origin is:
View full question & answer→MCQ 1231 Mark
The direction cosines of the normal to the plane $2x - 3y - 6z - 3 = 0$ are:
- ✓
$\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
- B
$\frac{2}{7},\frac{3}{7},\frac{6}{7}$
- C
$\frac{-2}{7},\frac{-3}{7},\frac{-6}{7}$
- D
AnswerCorrect option: A. $\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
View full question & answer→MCQ 1241 Mark
Cosine of the angle between two diagonals of acube is equal to:
- A
$\frac{2}{\sqrt{6}}$
- ✓
$\frac{1}{3}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: B. $\frac{1}{3}$
View full question & answer→MCQ 1251 Mark
The equation of the line passing through the points $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
- A
$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\lambda\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
- B
$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
- ✓
$\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
- D
$\text{None of these}$
AnswerCorrect option: C. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\texEquation of the line passing through the points having position vectors
$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$ where t is a parameter
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
View full question & answer→MCQ 1261 Mark
A parallelopiped is formed by planes drawn through the point (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes. The length of a diagonal of the parallelopiped is:
- ✓
$7$
- B
$\sqrt{38}$
- C
$\sqrt{155}$
- D
$\text{none of these}$
AnswerThe given point (2, 3, 5) and (5, 9, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates are different.
$\therefore$ Edges of the paralleloppiped
= |2 - 5|, |3 - 9| and |5 - 7|
=3, 6 and 2.
Now,
Length of the diagonal of the parallelopiped
$=\sqrt{3^2+6^2+2^2}$
$=\sqrt{9+36+4}$
$=\sqrt{49}$
$=7$
Hence, length of the diagonal of the parallelepiped formed by the planes
Parallel to coordinate planes and drawn through point (2, 3, 5)and (5, 9, 7) is 7 units.
View full question & answer→MCQ 1271 Mark
The distance between the planes $2x + 2y - z +2 = 0$ and $4x + 4y - 2z + 5 = 0$ is:
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- ✓
$\frac{1}{6}$
- D
AnswerCorrect option: C. $\frac{1}{6}$
Multiplying the first equation of the plane by
$4x + 4y - 2z + 4 = 0$
$4x + 4y - 2z = -4 .....(1)$
The second eqution of the plane is
$4x + 4y - 2z + 5 = 0$
$4x + 4y - 2z = -5 .....(2)$
We know that the distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is,
$=\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-5+4|}{\sqrt{4^2+4^2+(-2)^2}}$
$=\frac{|-1|}{\sqrt{16+16+4}}$
$=\frac{1}{\sqrt{36}}$
$=\frac{1}{6}\text{units}$
View full question & answer→MCQ 1281 Mark
The length of the perpendicular drawn from the point (4, -7, 3) on the y-axis is:
AnswerThe length of the perpendicular drawn from the point (4, -7, 3) on the y-axis is
⇒ Point on the y-axis would be = (0, -7, 0)
The length of the perpendicular drawn $=\sqrt{(4-0)^2}+(-7-(-7))^2+(3-0)^2$
$=\sqrt{4^2}+0^2+3^2$
$\Rightarrow\sqrt{16}+0+9$
$=\sqrt{25}$
$=5$
View full question & answer→MCQ 1291 Mark
A point P lies on the line segment joining the points (-1, 3, 2) and (5, 0, 6). If x-coordinate of P is 2, then its z-coordinate is:
- A
$-1$
- ✓
$4$
- C
$\frac{3}{2}$
- D
$8$
AnswerEquation of time passing through (-1, 3, 2) and (5, 0, 6)
$\frac{\text{x}+1}{5-(-1)}=\frac{\text{y}-3}{0-3}=\frac{\text{z}-2}{6-2}$
$\frac{\text{x}+1}{6}=\frac{\text{y}-3}{-3}$
$=\frac{\text{z}-2}{4}=\text{k}$
Any point on it,
$\text{P}(6\text{k}-1,\text{3}\text{k}+3,4\text{k}+2)$
x Coordinate $=2$
$=6\text{k}-1$
$\Rightarrow\text{k}=\text{y}_2$
z Coordinate $=4\text{k}+2$
$=4\Big(\frac{1}{2}\Big)+2$
$=2+2=4$
View full question & answer→MCQ 1301 Mark
The vector equation of the plane containing the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$ and the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ is:
- ✓
$\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
- B
$\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
- C
$\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
- D
$\text{None of these}$
AnswerCorrect option: A. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
Let the direction ratio of the required plane be proportinal to a, b, c.
Scince the required plane contains the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
It must pass through the point (-2, -3, 4) and it should be parallel to the line.
So, the equation of the plane is
a(x + 2) + b(y + 3) + c(z - 4) = 0 ....(1) and
3a - 2b - c = 0 ....(2)
It is given that plane (1) passes through the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ or (1, 2, 3).
a(1 + 2) + b(2 + 3) + c(3 - 4) = 0
3a + 5b - c = 0 .......(3)
So,
Solving (1) (2) and (3), we get
$\begin{vmatrix}\text{x}+2&\text{y}+3&\text{z}-4\\3&-2&-1\\3&5&-1\end{vmatrix}=0$
$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$
$\Rightarrow\text{x}+2+3\text{z}-12=0$
$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$
View full question & answer→MCQ 1311 Mark
The acute angle between the planes $2x - y + z = 0$ and $x + y + 2z = 3$ is:
- A
$45^\circ $
- ✓
$60^\circ $
- C
$30^\circ $
- D
$75^\circ $
AnswerCorrect option: B. $60^\circ $
We know that the angle between the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}$
So, the angle between $2x - y + z = 0$ and $x + y + 2x = 3$ is given by
So, $\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}$
$=\frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow\theta\cos^{-1}\Big(\frac{1}{2}\Big)=60^\circ$
View full question & answer→MCQ 1321 Mark
The eqution of the plane through the line $x + y + 3 = 0 = 2x - y + 3z + 1$ and parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ is:
- ✓
$x - 5y + 3z = 7$
- B
$x - 5y + 3z = -7$
- C
$x + 5y + 3z = 7$
- D
$x + 5y + 3z = -7$
AnswerCorrect option: A. $x - 5y + 3z = 7$
The equation of the plane passing though the line of intersection of the given planes is
$\text{x}+\text{y}+\text{z}+3+\lambda(2\text{x}-\text{y}+3\text{z}+1)=0$
$(1+2\lambda)\text{x}+(1-\lambda)\text{y}+(1+3\lambda)\text{z}+3+\lambda=0\ ....(1)$
This plane is parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}.$
It means that this line is perpendicular to the normal of the plane $(1).$
$\Rightarrow1(1+2\lambda)\text{x}+2(1-\lambda)+3(1+3\lambda)=0($Because $a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\Rightarrow1+2\lambda+2-2\lambda+3+9\lambda=0$
$\Rightarrow9\lambda+6=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-2}{3}\Big)\Big)\text{x}+\Big(1-\Big(\frac{-2}{3}\Big)\Big)\text{y}+\Big(1+3\Big(\frac{-2}{3}\Big)\Big)\text{z}+3+\Big(\frac{-2}{3}\Big)=0$
$\Rightarrow -x + 5y - 3z + 7 = 0$
$\Rightarrow x - 5y + 3z = 7$
View full question & answer→MCQ 1331 Mark
The coordinates of the midpoints of the line segment joining the points $(2, 3, 4)$ and $(8, -3, 8)$ are:
- A
$(10, 0, 12)$
- B
$(5, 6, 0)$
- C
$(6, 5, 0)$
- ✓
$(5, 0, 6)$
AnswerCorrect option: D. $(5, 0, 6)$
View full question & answer→MCQ 1341 Mark
The direction cosines of the $y-$axis are:
- A
$(9, 0, 0)$
- B
$(1, 0, 0)$
- ✓
$(0, 1, 0)$
- D
$(0, 0, 1)$
AnswerCorrect option: C. $(0, 1, 0)$
View full question & answer→MCQ 1351 Mark
The direction ratios of the line of intersection of the planes $3x + 2y - z = 5$ and $x - y + 2z = 3$ are:
- A
$3, 2, -1$
- ✓
$-3, 7, 5$
- C
$1, -1, 2$
- D
$-11, 4, -5$
AnswerCorrect option: B. $-3, 7, 5$
View full question & answer→MCQ 1361 Mark
lf a line makes angles $\frac{\pi}{12},\frac{5\pi}{12}$ with oy, oz respectively where 0 = (0, 0, 0), then the angle made by that line with ox is:
Answer$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{5\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1..$
$\Big(\cos\theta=\sin\Big(\frac{\pi}{2}- \theta\Big)\Big)$
$\Big(\cos(\gamma)\Big)^2=0$
$\cos(\gamma)=0$
$\gamma=90^\circ$
View full question & answer→MCQ 1371 Mark
The equation of the plane through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$ is:
AnswerCorrect option: C. $51x - 15y - 50z + 173 = 0$
The eqution of the plane passing through the line of intersection of the given planes is
$\text{x}+2\text{y}+3\text{z}-4+\lambda(2\text{x}+\text{y}-\text{z}+5)=0$
$(1+2\lambda)\text{x}+(2+\lambda)\text{y}+6(3-\lambda)\text{z}-4+5\lambda=0\ ....(1)$
This plane is perpendicular to $5x + 3y + 6z + 8 = 0.$ So,
$5(1+2\lambda)+3(2+\lambda)+6(3-\lambda)=0 ($Because $a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\Rightarrow5+10\lambda+6+3\lambda+18-6\lambda=0$
$\Rightarrow7\lambda+29=0$
$\Rightarrow\lambda=\frac{-29}{7}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-29}{7}\Big)\Big)\text{x}+\Big(2+\Big(\frac{-29}{7}\Big)\Big)\text{y}+\Big(3+\frac{29}{7}\Big)\text{z}-4+5\Big(\frac{-29}{7}\Big)=0$
$\Rightarrow 51x + 15y - 50z + 173 = 0.$
View full question & answer→MCQ 1381 Mark
Choose the correct answer from the given four options.
The plane 2x – 3y + 6z – 11 = 0 makes an angle $\sin^{-1}(\alpha)$ with x-axis. The value of $\alpha$ is equal to:
- A
$\frac{\sqrt{3}}{2}$
- B
$\frac{\sqrt{2}}{3}$
- ✓
$\frac{2}{7}$
- D
$\frac{3}{7}$
AnswerCorrect option: C. $\frac{2}{7}$
We are given that, 2x - 3y + 6z - 11 = 0 makes angle $\sin^{-1}(\alpha)$ with x-axis.
The equation of plane 2x - 3y + 6z - 11 = 0 in vector form is given by $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}})=11$
$\therefore\vec{\text{b}}=(\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$ and $\vec{\text{n}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
We know that, $\sin\theta=\frac{|\vec{\text{b}}\cdot\vec{\text{n}}|}{|\vec{\text{b}|}\cdot|\vec{\text{n}}|}$
$=\frac{\big|(\vec{\text{i}})\cdot(2\vec{\text{i}}-3\vec{\text{j}}+6\vec{\text{k}})\big|}{\sqrt{1}\sqrt{4+9+36}}$
$=\frac{2}{7}$
View full question & answer→MCQ 1391 Mark
The vector equation r = i − 2j − k + t(6j − k) represents a straight line passing through the points:
- A
(0, 6, −1) and (1, −2, −1)
- B
(0, 6, −1) and (−1, −4, −2)
- ✓
(1, −2, −1) and (1, 4, −2)
- D
(1, −2, −1) and (0, −6, 1)
AnswerCorrect option: C. (1, −2, −1) and (1, 4, −2)
Cartesian representation of the given line is,
$\frac{\text{x}-1}{0}=\frac{\text{y}+2}{6}=\frac{\text{z}+1}{-1}=\text{t}$
So any point on the given line is of the form (1, 6t − 2, − t − 1) where t can be any real numbers
So for t = 0 and 1 the corresponding points are (1, −2, −1) and (1, 4, −2)
You can check other options does not satisfy above point for any t.
View full question & answer→MCQ 1401 Mark
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
- A
$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
- B
$2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
- ✓
$-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
- D
$2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
AnswerCorrect option: C. $-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
Let the required vector be a $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\ ....(1)$
Since the vector is parallel to the line of intersection of the given planes,
3a - b + c = 0 .....(2)
a + 4b - 2c = 0 ....(3)
Solving (2) and (3), we get
$\frac{\text{a}}{-2}=\frac{\text{b}}{7}=\frac{\text{c}}{13}$
Substituting these values in (1), we get
$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$ which is the required vector.
View full question & answer→MCQ 1411 Mark
The distance of the point (-1, -5, -10) from the point of intersection of the line $\vec{\text{r}}.=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$ and the plane $\vec{\text{r}}.=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ is:
AnswerGiven equation of line is
$\vec{\text{r}}.=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$
$\vec{\text{r}}.=(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}$
The coordinates of any point on this line are of the from
$(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}$ or $(2+3\lambda,-1+4\lambda,2+12\lambda)$
Scince this point lies on the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5,$
$\Big[(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}\Big].(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5,$
$\Rightarrow2+3\lambda+1-4\lambda+2+12\lambda-5=0$
$\Rightarrow\lambda=0$
So, the coordinates of the point are
$(2+3\lambda,-1+4\lambda,2+2\lambda)$
$=(2+0,-1+0,2+0)$
$=(2, -1,2)$
Distance between (2, -1, 2) and (-1, -5, -10)
$=\sqrt{(1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13 \text{ units}$
View full question & answer→MCQ 1421 Mark
If $\alpha,\beta,\gamma$ are the angle which a half ray makes with the positive directions of the axis then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma =$
View full question & answer→MCQ 1431 Mark
The points $A(1, 1, 0), B(0, 1, 1), C(1, 0, 1)$ and $\text{D}\big(\frac{2}{3},\frac{2}{3},\frac{2}{3}\big)$
- ✓
- B
- C
Vertices of a parallelogram
- D
View full question & answer→MCQ 1441 Mark
If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$
AnswerIf a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then
$\cos2\alpha+\cos2\beta+\cos2\gamma=1\dots(1)$
We have
$\cos2\alpha+\cos2\beta+\cos2\gamma$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\gamma-1$ $\big[\because\cos2\theta=2\cos^2\theta-1\big]$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$ [From (1)]
$=2(1)-3$
$=-1$
View full question & answer→MCQ 1451 Mark
If the direction cosine of a directed line be $a, 3a, 7a$ then $a =$
- ✓
$\underline{+}\frac{1}{59}$
- B
$\underline{+}\frac{1}{9}$
- C
$\underline{+}\frac{2}{7}$
- D
AnswerCorrect option: A. $\underline{+}\frac{1}{59}$
Give$, a, 3a, 7a$ be the direction cosines of a directed line.
Then from the property of direction cosines we get
$a^2 + (3a)^2+ (7a)^2 = 1$ or
$59a^2 = 1$ or
$\text{a}=\underline{+}\frac{1}{\sqrt{59}}$
View full question & answer→MCQ 1461 Mark
The direction ratios of the line joining the points $(x, y, z)$ and $(x_2, y_2, z_1)$ are:
- A
$\text{x}_{1} + \text{x}_{2}, \text{y}_{1} +\text{ y}_{2}, \text{z}_{1} + \text{z}_{2}$
- B
$ (\text{x}_{1}-\text{x}_{2})^2+(\text{y}_{1}-\text{y}_{2})^2+(\text{z}_{1}+\text{z}_{2})^2$
- C
$\frac{\text{x}_{1}+\text{x}_{2}}{2}, \frac{\text{y}_{1}+\text{y}_{2}}{2}, \frac{\text{z}_{1}+\text{z}_{2}}{2}$
- ✓
$\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$
AnswerCorrect option: D. $\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$
View full question & answer→MCQ 1471 Mark
The equation of the plane through the origin and parallel to the plane $3x - 4y + 5z + 6 = 0:$
- A
$3x - 4y - 5z - 6 = 0$
- B
$3x - 4y + 5z + 6 = 0$
- ✓
$3x - 4y + 5z = 0$
- D
$3x + 4y - 5z + 6 = 0$
AnswerCorrect option: C. $3x - 4y + 5z = 0$
View full question & answer→MCQ 1481 Mark
A line makes an angle $\alpha,\beta,\gamma$ with the X, Y, Z axes. Then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$
AnswerFor a vector.
$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$
$1-\sin^2(\alpha)+1-\sin^2(\beta)+1-\sin^2(\gamma)=1$
$\sin^2(\alpha)\sin^2(\beta)+\sin^2(\gamma)=3-1=2$
View full question & answer→MCQ 1491 Mark
The Image of the point (2, -1, 5) in the plane $\vec{\text{r}},\hat{\text{i}}=0$ is:
AnswerEquation of plane is r.i = 0
i.e. x = 0
It is equation of Y-Z plane
Let PQ be the line Perpendicular to the plane from (2, -1, 5)
Also line is perpendicular to plane so direction ratios of line will be that of the DR's of plane equation of line will be:
$\frac{(\text{x}-2)}{(1)} = \frac{(\text{x}-b)}{0} = \frac{(\text{x}-\text{c})}{0} = \text{k say}$
General points of line PQ will be
x = k + 2
y = -1
z = 5
Also, this line intersect the plane
so, foot of perpendicular will be
(k + 2) = 0
k = -2
Hence foot of perpendicular will be (0, -1, 5)
let coordinates of image is (e, f, g)
By mid-point theorem
$0 = \frac{(2 + \text{e})}{2} \Rightarrow \text{e} = -2$
$-1 = \frac{(-1 + \text{f})}{2}\Rightarrow\text{f} = -1$
$5 = \frac{(5 + \text{g})}{2}\Rightarrow\text{g} = 5$
So, coordinates of image is
(-2, -1, 5)
View full question & answer→MCQ 1501 Mark
If the direction cosines of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$ then:
AnswerCorrect option: C. $\text{c}=\underline{+}\sqrt{2}$
Since, DC′s of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$
$\because\text{l}^2 + \text{m}^2 + \text{n}^2 = 1$
$\because\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2=1$
$\Rightarrow1 + 1 + 1 = \text{c}^2$
$\Rightarrow\text{c}^2 = 3$
$\Rightarrow\text{c}=\underline{+}\sqrt{3}$
View full question & answer→MCQ 1511 Mark
The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:
AnswerThe equations of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
Thus, the two lines are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and pass through the points (0, 0, 0) and (1, 2, 3).
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$
Since, the distence between the two parallel lines is 0, the given two lines are coincident lines.
Disclaimar: The answer given in the book is incorrect. This solution is created according to the question given in the book.
View full question & answer→MCQ 1521 Mark
The eqution of the plane which cute equal intercepts of unit length on the coordinate axes is:
AnswerWe know that the equation of aplane whose intercepts are a, b, c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(1)$
It is given that a = b = c
So, from (1),
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\text{x}+\text{y}+\text{z}=\text{a}\ ....(2)$
Since it is given that the intercepts of the required plane are of unit length,
a = b = c = 1
Substituting a = 1 in (2), we get
x + y + z = 1
View full question & answer→MCQ 1531 Mark
A straight line passes through (1, -2, 3) and perpendicular to the plane 2x + 3y - z = 7. Find the direction ratios of normal to plane:
Answerconcept: for any plane ax + by + cz + d =
0, normal vector to this plane is $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
the normal vector of the plane 2x + 3y - z = 7 is $\text{2}\hat{\text{i}}+\text{3}\hat{\text{j}}+\hat{\text{k}}$
so the direction ratios of normal to plane are < 2, 3, -1 >
View full question & answer→MCQ 1541 Mark
The direction coisines of the $y-$axis are:
- A
$(6, 0, 0)$
- B
$(1, 0, 0)$
- ✓
$(0, 1, 0)$
- D
$(0, 0, 1)$
AnswerCorrect option: C. $(0, 1, 0)$
View full question & answer→MCQ 1551 Mark
The distance of the point P(a, b, c) from the x-axis is:
- ✓
$\sqrt{\text{b}^2+\text{c}^2}$
- B
$\sqrt{\text{a}^2+\text{c}^2}$
- C
$\sqrt{\text{a}^2+\text{b}^2}$
- D
$\text{none of these}$
AnswerCorrect option: A. $\sqrt{\text{b}^2+\text{c}^2}$
The projection of the point P(a, b, c) on the x-axis is a, (0, 0) as both Y and Z coordinates on any point on the x-axis are equal to zero.
$\therefore$ Distance of P(a, b, c) from x-axis = Distance of P(a, b, c) from a, (0, 0)
$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$
$=\sqrt{\text{b}^2+\text{c}^2}$
View full question & answer→MCQ 1561 Mark
What are the DRs of vector parallel to (2, -1, 1) and (3, 4, -1):
AnswerRequired DRs are (3 - 2, 4 + 1, -1 - 1) ie, (1, 5, -2)
View full question & answer→MCQ 1571 Mark
If a plane passes through the point (1, 1, 1) and is perpendicular to the line $\frac{\text{x}-1}{3}=\frac{\text{y}-1}{0}=\frac{\text{z}-1}{4}$ then its perpendicular distance from the origin is:
- A
$\frac{3}{4}$
- B
$\frac{4}{3}$
- ✓
$\frac{7}{5}$
- D
$1$
AnswerCorrect option: C. $\frac{7}{5}$
Since the plane is perpendicular to the given line, its direction ratios are proportinal to 3, 0, 4.
So the required equation of the plane is of the form
3x + 0y + 4z + d = 0 .....(1), where d is a constant.
Since this plane passes through (1, 1, 1),
3 + 0 + 4 + d = 0
d = -7
Substituting this in (1), we get
3x + 0y + 4z -7 = 0 ......(2)
perpendicular distance of (2) from the origin
$=\frac{|3(0)+0+4(0)-7|}{\sqrt{3^2+0^2+4^2}}$
$=\frac{|0+0-7|}{\sqrt{25}}$
$=\frac{7}{5}\text{ units}$
View full question & answer→MCQ 1581 Mark
If l, m, n are the d.cs of the line joining (5, -3, 8) and (6, -1, 6) then l + m + n =
- A
$1$
- ✓
$\frac{1}{3}$
- C
$-1$
- D
$\frac{5}{3}$
AnswerCorrect option: B. $\frac{1}{3}$
The line joining (5, -3, 8) and (6, -1, 6) is given by the vector -i + 2j - 2k.
the direction cosines are given by. l =
$\frac{1}{\sqrt{1^2+2^2+(-2)^2}}=\frac{1}{3}, \text{m}=\frac{2}{\sqrt{1^2+2^2+(-2)^2}}=\frac{2}{3}$
$\text{n}=\frac{-2}{1^2+2^2+(-2)^2}=\frac{-2}{3}$
$\Rightarrow\text{l + m + n}=\frac{1}{3}$
View full question & answer→MCQ 1591 Mark
$l = m = n = 1$ represents the direction cosines of:
- A
$x−$axis
- B
$y−$axis
- C
$z−$axis
- ✓
AnswerSuppose$, l, m, n$ are direction cosines
$\Longrightarrow 1^2 + m^2 + n^2 = 1$
But $1 = m = n = 1$
$\Longrightarrow 3m^2 = 1$
$\Longrightarrow 1 = m = n = \frac{1}{\sqrt3}$
which are not direction cosines of either of the three $co-$ordinate axes.
View full question & answer→MCQ 1601 Mark
If a line makes angles $\alpha,\beta,\gamma,\delta$ with four diagonals of a cube, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$ is equal to:
- A
$\frac{1}{3}$
- B
$\frac{2}{3}$
- ✓
$\frac{4}{3}$
- D
$\frac{8}{3}$
AnswerCorrect option: C. $\frac{4}{3}$
Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR
The direction ratiosm of OP, AR, BS and CQ are
a - 0, a - 0, a - 0, i.e. a, a, a
0 - a, a - 0, a - 0, i.e. -a, a, a
a - 0, 0 - a, a - 0, i.e. a, -a, a
a - 0, a - 0, 0 - a, i.e. a, a, -a
Let the direction ratios of a line be proportional to l, m and n. Suppose this line makes angles $\alpha,\beta,\gamma$ and $\delta$ with OP, AR.
Now, $\alpha$ is the angle between OP and the line whose direction ratios are proportional to l, m and n.
$\cos\alpha=\frac{\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\alpha=\frac{\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Since $\beta$ is the angle between AR and the line with direction ratios proportional to l, m and n, we get
$\cos\beta=\frac{-\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\beta=\frac{-\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Similarly,
$\cos\gamma=\frac{\text{a}.\text{l}-\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\gamma=\frac{\text{l}-\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos\delta=\frac{\text{a}.\text{l}+\text{a}.\text{m}-\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\delta=\frac{\text{l}+\text{m}-\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$
$=\frac{(\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(-\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}-\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}+\text{m}-\text{n})^2}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}\Big\{(\text{l}+\text{m}+\text{n})^2+(-\text{l}+\text{m}+\text{n})^2+(\text{l}-\text{m}+\text{n})^2+(\text{l}+\text{m}-\text{n})^2\Big\}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}4\big(\text{l}^2+\text{m}^2+\text{n}^2\big)=\frac{4}{3}$. View full question & answer→MCQ 1611 Mark
The distance of the plane through the intersection of the planes ax + by + cz +d = 0 and lx + my + nz + P = 0 and parallel to the line y = 0, z = 0
- ✓
(bl - am)y + (cl - an)z + dl - ap = 0
- B
(am - bl)x + (mc - bn)z + md - bp = 0
- C
(na - cl)x + (bn - cm)y + nd - cp = 0
- D
AnswerCorrect option: A. (bl - am)y + (cl - an)z + dl - ap = 0
The equation of the plane passing through the intersection of the planes
ax + by + cz + d = 0
and lx + my + nz + p =0
Will be $(\text{ax} + \text{by} +\text{cz} +\text{d})+\lambda(\text{lm}+\text{my}+\text{nz}+\text{p})=0$
$\text{x}(\text{a}+\lambda1)+\text{y}(\text{b}+\lambda\text{m})+\text{z}(\text{c}+\lambda\text{n})+(\text{d}+\lambda\text{p})=0\ (1)$
Since the plane is parallel to the line y = 0 and z = 0
$\text{a}+\lambda1=0$
$\lambda=\frac{-\text{a}}{\text{l}}$
Putting the value of A in eqution (1), we get
$\text{x}\Big(\text{a}+\Big(\frac{\text{-a}}{\text{l}}\Big)\text{l}\Big)+\text{y}\Big(\text{b}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{m}+\text{y}\Big(\text{c}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{n}+\text{d}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{p}=0$
$\text{y}(\text{bl}-\text{am})+\text{z}(\text{cl}-\text{an})+\text{dl}-\text{ap}=0$
Heance, option (a)
View full question & answer→MCQ 1621 Mark
The eqution of the plane contaning the two lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$ is:
Answer$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$
Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.
View full question & answer→MCQ 1631 Mark
If a line makes the angle $\alpha,\beta,\gamma$ with three dimensional coordinate axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
AnswerWe need to find value of $\cos2\alpha+\cos2\beta+\cos2\gamma$
It is further equal to $\cos^2\alpha-1+\cos^2\beta-1+\cos^2\gamma-1$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$
$= 2(1) - 3 = 2 = -1$
$\therefore(\text{l}^2 + \text{m}^2 + \text{n}^2 = 1)$
View full question & answer→MCQ 1641 Mark
The perpendicular distance of the point P(1, 2, 3) from the line $\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$ is:
Answer$\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$
Let point (1, 2, 3) be P and the point through which the line passes be Q(6, 7, 7). Also, the line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Now,
$\overrightarrow{\text{PQ}}=5\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}} =\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&2&-2\\5&5&4\end{vmatrix}$
$=18\hat{\text{i}}-22\hat{\text{j}}+5\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{18^2+(-22)^2+5^2}$
$=\sqrt{324+484+25}$
$=\sqrt{833}$
$\because\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{833}}{\sqrt{17}}$
$=\sqrt{49}$
$=7$
View full question & answer→MCQ 1651 Mark
Can $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$ be the direction cosines of any directed line?
AnswerNo, they can not be the direction cosines of any directed line.
As the sum of square of them is not 1.
$\text{As}\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2$
$=\frac{1+4+4}{3}$
$=3$
View full question & answer→MCQ 1661 Mark
The angle between the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$ and $\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$ is:
AnswerCorrect option: C. $\frac{\pi}{3}$
We have
$\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$
$\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$
The direction ratios of the given lines are proportional to 1, 1, 2 and $-\sqrt{3}-1,\sqrt{3}-1, 4$
The given lines are parallel to vectors $\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big\{\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}}{\sqrt{1^2+1^2+1^2}\sqrt{\big(-\sqrt{3}-1\big)^2+\big(\sqrt{3}-1\big)^2+4^2}}$
$=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}}$
$=\frac{6}{6\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→MCQ 1671 Mark
(2, -3, -1) 2x - 3y + 6z + 7 = 0:
MCQ 1681 Mark
length of the $1^{er}$ from the point $(0, -1, 3)$ to the plane $2x + y - 2z + 1 = 0$ is:
- A
$0$
- B
$2\sqrt{3}$
- C
$\frac{2}{3}$
- ✓
$2$
View full question & answer→MCQ 1691 Mark
The projection of the join of the two points (1, 4, 5), (6, 7, 2) on the line whose d.ss are (4, 5, 6) is:
- ✓
$\frac{17}{\sqrt{77}}$
- B
$\frac{7}{6}$
- C
$21$
- D
$\frac{7}{9}$
AnswerCorrect option: A. $\frac{17}{\sqrt{77}}$
$\frac{17}{\sqrt{77}}$
View full question & answer→MCQ 1701 Mark
If the diraction ratios of a line are proportional to 1, -3, 2, then its diraction cosines are:
- ✓
$\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
- B
$\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
- C
$-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
- D
$-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}$
AnswerCorrect option: A. $\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
The diraction ratios of the line are proportional to 1, -3, 2.
$\therefore$ The direction cosines of the line are
$\frac{1}{\sqrt{1^2+(-3)^2+2^2}},\frac{-3}{\sqrt{1^2+(-3)^2+2^2}},\frac{2}{\sqrt{1^2+(-3)^2+2^2}}$
$=\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
View full question & answer→MCQ 1711 Mark
If a line makes angles $Q_1, Q_{21}$ and $Q_3$ respectively with the coordinate axis then the value of $\cos^2 \text{Q}_{1} + \cos^2 \text{Q}_{2} + \cos^2 \text{Q}_{3}:$
- A
$2$
- ✓
$1$
- C
$4$
- D
$\frac{3}{2}$
View full question & answer→MCQ 1721 Mark
The equation x² - x - 2 = 0 in three-dimensional space is represented by:
- ✓
A pair of parallel planes
- B
- C
A pair of the perpendicular plane
- D
AnswerCorrect option: A. A pair of parallel planes
A pair of parallel planes
View full question & answer→MCQ 1731 Mark
lf $\text{AB}\perp\text{BC}$ then the value of $\lambda$ equal, where A(2k, 2, 3), B(k, 1, 5), C(3 + k, 2, 1):
- A
$3$
- B
$\frac{1}{3}$
- ✓
$-3$
- D
$-\frac{1}{3}$
AnswerThe drs of AB are (k, 1, -2)
The drs of BC are (3, 1, -4)
Since, they are perpendicular, AB.BC = 0
3k + 1 + 8 = 0
k = -3
View full question & answer→MCQ 1741 Mark
If 2x + 5y - 6z + 3 = 0 be the equation of the plane, then the equation of any plane parallel to the given plane is:
MCQ 1751 Mark
The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:
AnswerWe have
$\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$
$\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$
The direction ratios of the given lines are proportional to 2, -3, 1 and 1, 2, -2.
The vectors parallel to the given vectors are $\vec{\text{b}}_1=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Vector perpendicular to the given two lines is
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&1\\1&2&-2\end{vmatrix}$
$=4\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$
Hence, the direction ration of the line perpendicular to the given two lines are proportional to 4, 5, 7.
View full question & answer→MCQ 1761 Mark
The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal to:
MCQ 1771 Mark
The equation of the plane passing through the points (3, 2, −1), (3, 4, 2) and (7, 0, 6) is 5x + 3y −2z = λ where λ is:
MCQ 1781 Mark
If the d.rs of two lines are 1, -2, 3 and 2, 0, 1, then the d.rs of the line perpendicular to both the given lines is:
AnswerOA and OB are given by (1, -2, 3), (2, 0, 1)
A line that will be perpendicular to both OA and OB can be obtained by doing the cross product of OA with OB.
Then, n = OA × OB
n = -2i + 5j + 4k
(-2, 5, 4).
View full question & answer→MCQ 1791 Mark
Can $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$ be the direction cosines of any directed line:
AnswerNo, they can not be the direction cosines of any directed line. As the sum of square of them is not 1.As
$=\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2=\frac{1+4+4}{3}=3$
View full question & answer→MCQ 1801 Mark
The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{3}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: C. $\frac{\pi}{3}$
$\frac{\pi}{3}$
View full question & answer→