Question 11 Mark
Find the angle between the lines $\text{2x = 3y = – z and 6x = – y = – 4z.}$
AnswerWriting standard form
$\frac{\text{x}}{3} =\frac{\text{y}}{2} = \frac{\text{z}}{-6}\ \text{and} \ {\frac{\text{x}}{2}} = \frac{\text{y}}{-12} = \frac{\text{z}}{-3}$
$\text{Finding } \theta = \frac{\pi}{2}$
View full question & answer→Question 21 Mark
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Answer$\cos^{2} 90^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$
$\cos \gamma = \pm\frac{\sqrt{3}}{2}, \gamma = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$
View full question & answer→Question 31 Mark
$\text{Find} \lambda, \text{if the vectors} \overrightarrow{\text{a}} = \hat{\text{i}} + 3\hat{\text{j}} + \hat{\text{k}}, \overrightarrow{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}}$ $\text{and} \overrightarrow{\text{c}} = \lambda \hat{\text{j}} + 3\hat{\text{k}} $ $\text{are coplanar}.$
Answer$\begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{vmatrix} = 0 \Rightarrow \lambda = 7 $
View full question & answer→Question 41 Mark
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
Answer$\frac{\text{x}}{\sqrt{3}} + \frac{\text{y}}{\sqrt{3}} + \frac{\text{z}}{\sqrt{3}} = 5\sqrt{3} \text{ or } \text{x + y + z = 15}$
View full question & answer→Question 51 Mark
Write the vector equation of a line given by $\frac{\text{x - 5}}{3}=\frac{\text{y + 4}}{7}=\frac{\text{z - 6}}{2}.$
Answer$\vec{\text{r}}=(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}})+\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}})$.
View full question & answer→Question 61 Mark
Find the length of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 21 = 0.
AnswerGiven plane is
2x - 3y + 6z + 21 = 0
$\therefore\text{ Length of }\bot \text{ar from origin (0, 0, 0)} = \bigg|\frac{0\times2 + 0 \times(-3) + 0 \times6+21}{\sqrt{2^{2} + (-3)^{2} + 6 ^{2}}}\bigg|$
$ = \bigg|\frac{21}{\sqrt{4 + 9 + 36}}\bigg| = \frac{21}{\sqrt{49}} = \frac{21}{7} = 3 $
Note: If $\rho$ is perpendicular distance from ($\alpha,\beta,\gamma$) to plane ax + by + cz + d = 0 then
$ =\text{p} = \bigg|\frac{\text{a}\alpha + \text{b}\beta +\text{c}\gamma +\text{d}}{\sqrt{\text{a}^{2} + \text{b}^{2} + \text{c}^{2}}}\bigg|.$
View full question & answer→Question 71 Mark
Write the distance of the point (3, – 5, 12) from x-axis.
Answer$\sqrt{(-5)^{2} + (12)^{2}} = 13$
View full question & answer→Question 81 Mark
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
Answer$\frac{\text{x}}{\sqrt{3}} + \frac{\text{y}}{\sqrt{3}} + \frac{\text{z}}{\sqrt{3}} = 5\sqrt{3} \text{ or } \text{x + y + z = 15}$
View full question & answer→Question 91 Mark
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Answer$\cos^{2} 90^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$
$\cos \gamma = \pm\frac{\sqrt{3}}{2}, \gamma = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$
View full question & answer→Question 101 Mark
Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
Answer$-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.
View full question & answer→Question 111 Mark
Write tbe distance of the following plane from the origin: 2x – y + 2z + 1 = 0
Question 121 Mark
Find the value of ‘p’ for which the vectors $3\hat{\text{i}} + 2\hat{\text{j}} + 9 \hat{\text{k}}\text{ and } \hat{\text{i}} - 2\text{p} \hat{\text{j}} + 3\hat{\text{k}}\text{ are parallel. }$
Answer$\text{p} = -\frac{1}{3}.$
View full question & answer→Question 131 Mark
Find the angle between the lines $\text{2x = 3y = – z and 6x = – y = – 4z.}$
AnswerWriting standard form
$\frac{\text{x}}{3} =\frac{\text{y}}{2} = \frac{\text{z}}{-6}\ \text{and} \ {\frac{\text{x}}{2}} = \frac{\text{y}}{-12} = \frac{\text{z}}{-3}$
$\text{Finding } \theta = \frac{\pi}{2}$
View full question & answer→Question 141 Mark
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
Answer$\frac{\text{x}}{\sqrt{3}} + \frac{\text{y}}{\sqrt{3}} + \frac{\text{z}}{\sqrt{3}} = 5\sqrt{3} \text{ or } \text{x + y + z = 15}$
View full question & answer→Question 151 Mark
Write the distance of the point (3, – 5, 12) from x-axis.
Question 161 Mark
If a line has direction ratios 2, –1, –2, then what are its direction cosines?
Answer$\frac{2}{3},\frac{-1}{3},\frac{-2}{3}.$.
View full question & answer→Question 171 Mark
Find the distance between the planes $2x – y + 2z = 5$ and $5x – 2.5y + 5z = 20$.
AnswerConsider the equations of planes, 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20.
Here, we can see the above two planes are parallel planes.
As, 5x – 2.5y + 5z = 20 can also be written as 2x – y + 2z = 8
If the equation of two parallel planes are
$ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$
Then, distance between the two parallel planes is given by:
$d = \bigg|\frac{(d_{2} - d_{1})}{\sqrt{a^{2} + b^{2} + c^{2}}}\bigg|$
let us take the two parallel planes be 2x – y + 2z = 5 and 2x – y + 2z = 8
Therefore the distance is given by:
$d = \bigg|\frac{(5 - 8)}{\sqrt{2^{2} + (-1)^{2} + 2^{2}}}\bigg|$
$ = \bigg|\frac{-3}{\sqrt{4 + 1 + 4}}\bigg|$
$= \big|\frac{-3}{3}\big|$
= 1
Hence the distance between the given two planes is 1 units.
View full question & answer→Question 181 Mark
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Answer$\cos^{2} 90^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$
$\cos \gamma = \pm\frac{\sqrt{3}}{2}, \gamma = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$
View full question & answer→Question 191 Mark
Find the vector equation of the plane with intercepts $\text{3, -4 and 2 on } x, y \text{ and } z-\text{axis}$ respectively.
Answer$\frac{\text{x}}{3} + \frac{\text{y}}{-4} + \frac{\text{z}}{2} = 1$
$\Rightarrow\overrightarrow{\text{r}}.(4\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}) = 12 \text{ or} \overrightarrow{\text{r}}. \bigg(\frac{\hat{\text{i}}}{3} - \frac{\hat{\text{j}}}{4} + \frac{\hat{\text{k}}}{2}\bigg) = 1$
View full question & answer→Question 201 Mark
Find the distance of the plane 3x – 4y + 12z = 3 from the origin.
AnswerDistance of plane from origin $=\frac{\text{D}}{\sqrt{\text{A}^2+\text{B}^2+\text{C}^2}}$
$=\frac{3}{\sqrt{9+16+144}}$
$=\frac{3}{\sqrt{169}}$
$\frac{3}{13}.$
View full question & answer→Question 211 Mark
$\text{Find} \lambda, \text{if the vectors} \overrightarrow{\text{a}} = \hat{\text{i}} + 3\hat{\text{j}} + \hat{\text{k}}, \overrightarrow{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}}$ $\text{and} \overrightarrow{\text{c}} = \lambda \hat{\text{j}} + 3\hat{\text{k}} $ $\text{are coplanar}.$
Answer$\begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{vmatrix} = 0 \Rightarrow \lambda = 7 $
View full question & answer→Question 221 Mark
Write the direction cosines of a line equally inclined to the three coordinate axes.
Answer$\Bigg[\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Bigg].$
View full question & answer→Question 231 Mark
$\text{Find}\ \lambda\ \text{if}\ (2\hat{\text{i}}+6\hat{\text{j}}+14\hat{\text{k}})\times(\hat{\text{i}}-\lambda\hat{\text{i}}+7\hat{\text{k}})=\overrightarrow{0}\dot{}$
View full question & answer→Question 241 Mark
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\overrightarrow{\text{r}}.(\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}) = 2.$
Answer$\left\{\overrightarrow{\text{r}} - \big(\text{a}\hat{\text{i}} + \text{b}\hat{\text{j}} + \text{c}\hat{\text{k}}\big)\right\}.\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) = 0$
Alternate Answer
$\overrightarrow{\text{r}}\cdot\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) =\text{a } + \text{b } +\text{c}.$
View full question & answer→Question 251 Mark
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\overrightarrow{\text{r}}.(\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}) = 2.$
Answer$\left\{\overrightarrow{\text{r}} - \big(\text{a}\hat{\text{i}} + \text{b}\hat{\text{j}} + \text{c}\hat{\text{k}}\big)\right\}.\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) = 0$
Alternate Answer
$\overrightarrow{\text{r}}\cdot\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) =\text{a } + \text{b } +\text{c}.$
View full question & answer→Question 261 Mark
Find the distance between the planes $2x – y + 2z = 5$ and $5x – 2.5y + 5z = 20.$
AnswerConsider the equations of planes, $2x – y + 2z = 5$ and $5x – 2.5y + 5z = 20.$
Here, we can see the above two planes are parallel planes.
As, $5x – 2.5y + 5z = 20$ can also be written as $2x – y + 2z = 8$
If the equation of two parallel planes are
$ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$
Then, distance between the two parallel planes is given by:
$d = \bigg|\frac{(d_{2} - d_{1})}{\sqrt{a^{2} + b^{2} + c^{2}}}\bigg|$
let us take the two parallel planes be $2x – y + 2z = 5$ and $2x – y + 2z = 8$
Therefore the distance is given by:
$d = \bigg|\frac{(5 - 8)}{\sqrt{2^{2} + (-1)^{2} + 2^{2}}}\bigg|$
$ = \bigg|\frac{-3}{\sqrt{4 + 1 + 4}}\bigg|$
$= \big|\frac{-3}{3}\big|$
$= 1$
Hence the distance between the given two planes is $1$ units.
View full question & answer→Question 271 Mark
Find the vector equation of the plane with intercepts $\text{3, -4 and 2 on } x, y \text{ and } z-\text{axis}$ respectively.
Answer$\frac{\text{x}}{3} + \frac{\text{y}}{-4} + \frac{\text{z}}{2} = 1$
$\Rightarrow\overrightarrow{\text{r}}.(4\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}) = 12 \text{ or} \overrightarrow{\text{r}}. \bigg(\frac{\hat{\text{i}}}{3} - \frac{\hat{\text{j}}}{4} + \frac{\hat{\text{k}}}{2}\bigg) = 1$
View full question & answer→Question 281 Mark
In a triangle OAC, if B is the mid - point of side AC and $\overrightarrow{\text{OA}} = \overrightarrow{\text{a}}, \overrightarrow{\text{OB}} = \overrightarrow{\text{b}}, $ then what is $\overrightarrow{\text{OC}} \text{?}$
Answer$\overrightarrow{\text{OB}} = \frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}{2}$
$\overrightarrow{\text{OC}} = \overrightarrow{2\text{b}} - \overrightarrow{\text{a}}$
View full question & answer→Question 291 Mark
If the cartesian equations of a line are$\frac{3 - \text{x}}{5} = \frac{\text{y} + 4 }{7} = \frac{2\text{z} - 6 }{4},$ write the vector equation for the line.
Answer$\overrightarrow{\text{r}} = ( 3 \hat{\text{i}} -4\hat{\text{j}} + 3 \hat{\text{k}}) + \lambda( - 5\hat{\text{i}} + 7 \hat{\text{j}} + 2 \hat{\text{k}}).$
View full question & answer→Question 301 Mark
In a triangle OAC, if B is the mid - point of side AC and $\overrightarrow{\text{OA}} = \overrightarrow{\text{a}}, \overrightarrow{\text{OB}} = \overrightarrow{\text{b}}, $ then what is $\overrightarrow{\text{OC}} \text{?}$
Answer$\overrightarrow{\text{OB}} = \frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}{2}$
$\overrightarrow{\text{OC}} = \overrightarrow{2\text{b}} - \overrightarrow{\text{a}}$
View full question & answer→Question 311 Mark
Find the value of ‘p’ for which the vectors $3\hat{\text{i}} + 2\hat{\text{j}} + 9 \hat{\text{k}}\ \text{ and } \ \hat{\text{i}} - 2\text{p} \hat{\text{j}} + 3\hat{\text{k}}\text{ are parallel. }$
Answer$\text{p} = -\frac{1}{3}.$
View full question & answer→Question 321 Mark
Find the value of ‘p’ for which the vectors $3\hat{\text{i}} + 2\hat{\text{j}} + 9 \hat{\text{k}}a\text{ and } \hat{\text{i}} - 2\text{p} \hat{\text{j}} + 3\hat{\text{k}}\text{ are parallel. }$
Answer$\text{p} = -\frac{1}{3}.$
View full question & answer→Question 331 Mark
If the cartesian equations of a line are$\frac{3 - \text{x}}{5} = \frac{\text{y} + 4 }{7} = \frac{2\text{z} - 6 }{4},$ write the vector equation for the line.
Answer$\overrightarrow{\text{r}} = ( 3 \hat{\text{i}} -4\hat{\text{j}} + 3 \hat{\text{k}}) + \lambda( - 5\hat{\text{i}} + 7 \hat{\text{j}} + 2 \hat{\text{k}}).$
View full question & answer→Question 341 Mark
Find the vector equation of the plane with intercepts $\text{3, -4 and 2 on } x, y \text{ and } z-\text{axis}$ respectively.
Answer$\frac{\text{x}}{3} + \frac{\text{y}}{-4} + \frac{\text{z}}{2} = 1$
$\Rightarrow\overrightarrow{\text{r}}.(4\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}) = 12 \text{ or} \overrightarrow{\text{r}}. \bigg(\frac{\hat{\text{i}}}{3} - \frac{\hat{\text{j}}}{4} + \frac{\hat{\text{k}}}{2}\bigg) = 1$
View full question & answer→Question 351 Mark
In a triangle OAC, if B is the mid - point of side AC and $\overrightarrow{\text{OA}} = \overrightarrow{\text{a}}, \overrightarrow{\text{OB}} = \overrightarrow{\text{b}}, $ then what is $\overrightarrow{\text{OC}} \text{?}$
Answer$\overrightarrow{\text{OB}} = \frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}{2}$
$\overrightarrow{\text{OC}} = \overrightarrow{2\text{b}} - \overrightarrow{\text{a}}$
View full question & answer→Question 361 Mark
Write the direction cosines of the vector $-\hat{\text{2i}}+\hat{\text{j}}-\hat{\text{5k}}$.
Answer$-\frac{{2}}{\sqrt{30}}, \frac{{1}}{\sqrt{30}},-\frac{{-5}}{\sqrt{30}}$.
View full question & answer→Question 371 Mark
Find the distance between the planes $2x – y + 2z = 5$ and $5x – 2.5y + 5z = 20$.
AnswerConsider the equations of planes, 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20.
Here, we can see the above two planes are parallel planes.
As, 5x – 2.5y + 5z = 20 can also be written as 2x – y + 2z = 8
If the equation of two parallel planes are
$ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$
Then, distance between the two parallel planes is given by:
$d = \bigg|\frac{(d_{2} - d_{1})}{\sqrt{a^{2} + b^{2} + c^{2}}}\bigg|$
let us take the two parallel planes be 2x – y + 2z = 5 and 2x – y + 2z = 8
Therefore the distance is given by:
$d = \bigg|\frac{(5 - 8)}{\sqrt{2^{2} + (-1)^{2} + 2^{2}}}\bigg|$
$ = \bigg|\frac{-3}{\sqrt{4 + 1 + 4}}\bigg|$
$= \big|\frac{-3}{3}\big|$
= 1
Hence the distance between the given two planes is 1 units.
View full question & answer→Question 381 Mark
Write the distance of the point (3, – 5, 12) from x-axis.
Answer$\sqrt{(-5)^{2} + (12)^{2}} = 13$
View full question & answer→Question 391 Mark
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\overrightarrow{\text{r}}.(\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}) = 2.$
Answer$\left\{\overrightarrow{\text{r}} - \big(\text{a}\hat{\text{i}} + \text{b}\hat{\text{j}} + \text{c}\hat{\text{k}}\big)\right\}.\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) = 0$
Alternate Answer
$\overrightarrow{\text{r}}\cdot\big(\hat{\text{i}} + \hat{\text{j}} +\hat{\text{k}}\big) =\text{a } + \text{b } +\text{c}.$
View full question & answer→Question 401 Mark
Find the angle between the lines $\text{2x = 3y = – z and 6x = – y = – 4z.}$
AnswerWriting standard form
$\frac{\text{x}}{3} =\frac{\text{y}}{2} = \frac{\text{z}}{-6}\ \text{and} \ {\frac{\text{x}}{2}} = \frac{\text{y}}{-12} = \frac{\text{z}}{-3}$
$\text{Finding } \theta = \frac{\pi}{2}$
View full question & answer→Question 411 Mark
Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is parallel to the line $\frac{\text{x} + 3 }{3} =\frac{4 - \text{y}}{5} = \frac{\text{z} + 8}{6}.$
AnswerEquation of required line is
$\frac{\text{x} - (-2)}{3} =\frac{\text{y} - 4 }{-5} = \frac{\text{z} - ( - 5 )}{6}$ [$\because$Direction ratios of given line are 3, – 5, 6]
$\Rightarrow\frac{\text{x} + 2 }{3} =\frac{\text{y} - 4 }{-5} =\frac{\text{z} + 5}{ 6}.$
View full question & answer→Question 421 Mark
Find $\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c}}) , \text{ if } \overrightarrow{\text{a}} = 2\hat{\text{i}} + \hat{\text{j}} + 3 \hat{\text{k}} , \overrightarrow{\text{b}} = - \hat{\text{i}} + 2 \hat{\text{j} + \hat{\text{k}}} \text{ and } \overrightarrow{\text{c}} = 3 \hat{\text{i}} + \hat{\text{j}} + 2 \hat{\text{k}}.$
View full question & answer→Question 431 Mark
If the cartesian equations of a line are$\frac{3 - \text{x}}{5} = \frac{\text{y} + 4 }{7} = \frac{2\text{z} - 6 }{4},$ write the vector equation for the line.
Answer$\overrightarrow{\text{r}} = ( 3 \hat{\text{i}} -4\hat{\text{j}} + 3 \hat{\text{k}}) + \lambda( - 5\hat{\text{i}} + 7 \hat{\text{j}} + 2 \hat{\text{k}}).$
View full question & answer→Question 441 Mark
Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.
Question 451 Mark
$\text{Find} \lambda, \text{if the vectors} \overrightarrow{\text{a}} = \hat{\text{i}} + 3\hat{\text{j}} + \hat{\text{k}}, \overrightarrow{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}}$ $\text{and} \overrightarrow{\text{c}} = \lambda \hat{\text{j}} + 3\hat{\text{k}} $ $\text{are coplanar}.$
Answer$\begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{vmatrix} = 0 \Rightarrow \lambda = 7 $
View full question & answer→