MCQ 1011 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 1021 Mark
A plane passing through $(−1, 2, 3)$ and whose normal makes equal angle with the coordinate axes is:
- A
$x + y + z + 4 = 0$
- B
$x − y + z + 4 = 0$
- ✓
$x + y + z − 4 = 0$
- D
$x + y + z = 0$
AnswerCorrect option: C. $x + y + z − 4 = 0$
Since normal makes equal angles with coordinate axis.
So, it intercept with all the axis will be same.
So equation of plane will be
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}+\frac{\text{x}}{\text{a}}=1$
$\Rightarrow x + y + z = a$
Now, it passes through $(−1, 2, 3),$ so
$−1 + 2 + 3 = a$
$\Rightarrow a = 4$
$\Rightarrow x + y + z − 4 = 0$
View full question & answer→MCQ 1031 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 1041 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:
- ✓
$5x + y − 11 = 0$
- B
$5x + 13y + 5= 0$
- C
$5x − 13y + 5 = 0$
- D
$13x − 5y + 5 = 0$
AnswerCorrect option: A. $5x + y − 11 = 0$
Since, the points $A(1, 6), B(3, −4)$ and $C(x, y)$ are colinear
$\therefore\begin{vmatrix}1 6 13 -4 1\text{x} \text{y} 1\end{vmatrix}=0$
$\Rightarrow 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0$
$\Rightarrow 10x + 2y − 22 = 0$
$\Rightarrow 5x + y − 11 = 0$
View full question & answer→MCQ 1051 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 1061 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 1071 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 1081 Mark
A line makes angles $\alpha,\beta,\gamma$ with the positive direction of the axes of reference. The value of $\cos^2\alpha+\cos^2\beta+\cos^2\gamma$ is:
Answer$\cos^2\alpha+\cos^2\beta+\cos^2\text{r}$
$=1\cos^2\alpha+\cos^2\beta+\cos^2\text{r}$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\text{r}-1$
$=2(\cos^2\alpha+\cos^2\beta\cos{\text{r}})-3$
$=2(1)-3$
$=-1$
View full question & answer→MCQ 1091 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
$\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1\ [$From $(1)]$
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=3$
View full question & answer→MCQ 1101 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 1111 Mark
If the projections of the line segment $AB$ on the coordinate axes are $12, 3, k$ and $AB = 13$ then $k^2-2 k+3$ 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 $k ^2-2 k +3=11$ or $27 .$
View full question & answer→MCQ 1121 Mark
What are the $DR\ 's$ of vector parallel to $(2, −1, 1)$ and $(3, 4, −1)\ ?$
- ✓
$(1, 5, −2)$
- B
$(−2, −5, 2)$
- C
$(−1, 5, 2)$
- D
$(−1, −5, −2)$
AnswerCorrect option: A. $(1, 5, −2)$
Required $DR\ 's$ are $(3 − 2, 4 + 1, −1−1)$ ie$, (1, 5, −2)$
View full question & answer→MCQ 1131 Mark
Choose the correct answer from the given four options.The area of the quadrilateral $\text{ABCD},$ where $A(0, 4, 1), B(2, 3, -1), C(4, 5, 0)$ and $D(2, 6, 2),$ is equal to:
- ✓
$\text{9 sq. units.}$
- B
$\text{18 sq. units.}$
- C
$\text{27 sq. units.}$
- D
$\text{81 sq. units.}$
AnswerCorrect option: A. $\text{9 sq. units.}$
We 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 $\text{ABCD}$
$=\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{BC}}\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-1&-2\\2&2&1 \end{vmatrix}$
$=|3\vec{\text{i}}-6\vec{\text{j}}+6\vec{\text{k}}|$
$=\sqrt{9+36+36}$
$=9\text{ sq.units}$
View full question & answer→MCQ 1141 Mark
Area of $\triangle\text{ABC}$ is:
- A
$45$ squares units
- B
$55$ squares units
- C
$65$ squares units
- ✓
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
$\text{PABC}=\frac{125}{8}$ cubic units
View full question & answer→MCQ 1151 Mark
The points with position vectors $60i + 3j, 40i − 8j$ and $ai − 52j$ are collinear if:
- ✓
$a = −40$
- B
$a = 40$
- C
$a = 20$
- D
AnswerCorrect option: A. $a = −40$
Denoting $a,b,c$ by the given vectors respectively
These vectors will be collinear if there is some constant $k$
such that $c − a = K(b − a)$
$\Rightarrow a − 60 = −20K$ and $−55 = −11K$
$\Rightarrow a = −100 + 60 = −40$
View full question & answer→MCQ 1161 Mark
The angle made by line $\text{r}[\cos\theta−3\sin\theta]=5$ with initial line is:
- ✓
$30^\circ $
- B
$45^\circ $
- C
$60^\circ $
- D
$90^\circ $
AnswerCorrect option: A. $30^\circ $
Given 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 $30^\circ$ with the initial line.
View full question & answer→MCQ 1171 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 1181 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 1191 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 1201 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 1211 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
AnswerCorrect option: C. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
Equation 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 1221 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
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 1231 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 $a x+b y+c z=d_1$ and $a x+b y+c z=d_2$ 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 1241 Mark
The length of the perpendicular drawn from the point $(4, -7, 3)$ on the $y-$axis is:
- A
$3$ units
- B
$4$ units
- ✓
$5$ units
- D
$7$ units
AnswerCorrect option: C. $5$ units
The length of the perpendicular drawn from the point $(4, -7, 3)$ on the $y-$axis is
$\Rightarrow$ 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 1251 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 1261 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
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
$\begin{vmatrix}\text{x}+2 \text{y}+3 \text{z}-43 -2 -13 5 -1\end{vmatrix}=0$
$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$
$\Rightarrow\text{x}+2+3\text{z}-12=0$
$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$
View full question & answer→MCQ 1271 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 $a_1 x+b_1 y+c_1 z+d_1=0$ and $a_2 x+b_2 y+c_2 z+d_2=0$ 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 1281 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 $\left.a_1 a_2+b_1 b_2+c_1 c_2=0\right)$
$\Rightarrow1+2\lambda+2-2\lambda+3+9\lambda=0$
$\Rightarrow9\lambda+6=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-2}{3}\Big)\Big)\text{x}+\Big(1-\Big(\frac{-2}{3}\Big)\Big)\text{y}+\Big(1+3\Big(\frac{-2}{3}\Big)\Big)\text{z}+3+\Big(\frac{-2}{3}\Big)=0$
$\Rightarrow-x+5 y-3 z+7=0$
$\Rightarrow x-5 y+3 z=7$
View full question & answer→MCQ 1291 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 1301 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 1311 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 1321 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:
- A
$45^\circ$
- ✓
$90^\circ$
- C
$60^\circ$
- D
$30^\circ$
AnswerCorrect option: B. $90^\circ$
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{5\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$
$\Big(\cos\theta=\sin\Big(\frac{\pi}{2}- \theta\Big)\Big)$
$\Big(\cos(\gamma)\Big)^2=0$
$\cos(\gamma)=0$
$\gamma=90^\circ$
View full question & answer→MCQ 1331 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,
$5(1+2\lambda)+3(2+\lambda)+6(3-\lambda)=0$ $\left(\right.$ Because $\left.a_1 a_2+b_1 b_2+c_1 c_2=0\right)$
$\Rightarrow5+10\lambda+6+3\lambda+18-6\lambda=0$
$\Rightarrow7\lambda+29=0$
$\Rightarrow\lambda=\frac{-29}{7}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-29}{7}\Big)\Big)\text{x}+\Big(2+\Big(\frac{-29}{7}\Big)\Big)\text{y}+\Big(3+\frac{29}{7}\Big)\text{z}-4+5\Big(\frac{-29}{7}\Big)=0$
$\Rightarrow 51 x+15 y-50 z+173=0$
View full question & answer→MCQ 1341 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 1351 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 1361 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 1371 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 $ units
View full question & answer→MCQ 1381 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 1391 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
Non$-$coplanar
- C
Vertices of a parallelogram
- D
View full question & answer→MCQ 1401 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 1411 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
$a^2+(3 a)^2+(7 a)^2=1$ or
$59 a^2=1$ or
$\text{a}=\underline{+}\frac{1}{\sqrt{59}}$
View full question & answer→MCQ 1421 Mark
The direction ratios of the line joining the points $(x, y, z)$ and $\left( x _2, y _2, z _1\right)$ 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 1431 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 1441 Mark
A line makes an angle $\alpha,\beta,\gamma$ with the $X, Y, Z$ axes. Then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$
- A
$1$
- ✓
$2$
- C
$\dfrac{3}{2}$
- D
$4$
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 1451 Mark
The Image of the point $(2, -1, 5)$ in the plane $\vec{\text{r}},\hat{\text{i}}=0$ is :
- ✓
$(-2, -1, 5)$
- B
$(2, 1, -5)$
- C
$(-2, 1, -5)$
- D
$(2, 0, 0)$
AnswerCorrect option: A. $(-2, -1, 5)$
Equation of plane is $\ce{r.i} = 0$
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 $\text{DR's}$ 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 $\text{(e, f, g)}$
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 1461 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 1471 Mark
The eqution of the plane which cute equal intercepts of unit length on the coordinate axes is:
- ✓
$x + y + z = 1$
- B
$x + y + z = 0$
- C
$x + y - z = 1$
- D
$x + y + z = 2$
AnswerCorrect option: A. $x + y + z = 1$
We 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 1481 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:
- ✓
$< 2, 3, -1 >$
- B
$< 2, 3, 1 >$
- C
$< -1, 2, 3 >$
- D
AnswerCorrect option: A. $< 2, 3, -1 >$
concept: 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 1491 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 1501 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
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→