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→