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→