Question 13 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=6$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})=9$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{n}}_2=3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}\big|}$
$=\frac{6-6-4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{-4}{(3)(7)}$
$=\frac{-4}{21}$
$\theta=\cos^{-1}\Big(\frac{-4}{21}\Big)$
View full question & answer→Question 23 Marks
Find the distance of the plane 2x- 3y + 4z - 6 = 0 from the origin.
AnswerThe given equation of the plane is,
2x - 3y + 4z = 6 .....(i)
Now, $\sqrt{2^2+(-3)^2+4^2}$
$=\sqrt{4+9+16}$
$=\sqrt{29}$
Dividing (i) by $\sqrt{29},$ we get
$\frac{2}{\sqrt{29}}\text{x}-\frac{3}{\sqrt{29}}\text{y}+\frac{4}{\sqrt{29}}\text{z}=\frac{6}{\sqrt{29}},$ which is the normal form of plane (i)
So, the length of the perpendicular from the origin to the plane $=\frac{6}{\sqrt{29}}$
View full question & answer→Question 33 Marks
Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z - 9 = 0$.
AnswerWe know that the distance of the point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by
$\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|(2)+2(3)-2(-5)-9|}{\sqrt{1^2+2^2+(-2)^2}}$
$=\frac{|2+6+10-9|}{\sqrt{1+4+4}}$
$=\frac{9}{3}$
$=3\text{ units}$
View full question & answer→Question 43 Marks
Find the equation of the plane through the untersection of the planes 3x - y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
AnswerThe equation of the family of planes through the intersection of planes 3x - y + 2z = 4 and x + y + z = 2 is,
$(3\text{x}-\text{y}+2\text{z}-4)+\lambda(\text{x}+\text{y}+\text{z}-2)=0\ ....(\text{i})$
If it passes through (2, 2, 1), then
$(6-2+2-4)+\lambda(2+2+1-2)=0$
$\lambda=-\frac{2}{3}$
Substituting $\lambda=-\frac{2}{3}$ in (i) we get, 7x - 5y + 4z = 0 as the equation of the required plane.
View full question & answer→Question 53 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
4x + 3y - 6z - 12 = 0
AnswerEquation of the given plane is,
4x + 3y - 6z - 12 = 0
⇒ 4x + 3y - 6z = 12
Dividng both sides by 12, we get
$\frac{4\text{x}}{12}+\frac{3\text{y}}{12}+\frac{(6\text{z})}{12}=\frac{12}{12}$
$\Rightarrow\frac{4\text{x}}{12}+\frac{3\text{y}}{12}-\frac{6\text{z}}{12}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{4}+\frac{\text{z}}{-2}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=3;\text{ b}=4;\text{ c}=-2$
View full question & answer→Question 63 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$3\text{x}-6\text{y}-2\text{z}=7$ and $2\text{x}+\text{y}-\lambda\text{z}=5$
AnswerWe know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are 3x - 6y - 2z = 7 and $2\text{x}+\text{y}-\lambda\text{z}=5$
$\Rightarrow a_1 = 3; b_1 = -6; c_1 = -2; a_2 = 2; b_2 = 1$; $\text{c}_2=-\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(3)(2)+(-6)(1)+(-2)(-\lambda)=0$
$\Rightarrow6-6+2\lambda=0$
$\Rightarrow2\lambda=0$
$\Rightarrow\lambda=0$
View full question & answer→Question 73 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
yz-plane
AnswerDirection ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the yz-plane x = 0
$\Rightarrow2\text{r}+5=0$
$\Rightarrow\text{r}=-\frac{5}{2}$
$\Rightarrow\text{y}=-3\Big(-\frac{5}{2}\Big)+1=\frac{17}{2}$
$\Rightarrow\text{z}=5\Big(-\frac{5}{2}\Big)+6=-\frac{13}{2}$
Hence, the corrdinates of the point are $\Big(0,\frac{17}{2},-\frac{13}{2}\Big)$
View full question & answer→Question 83 Marks
Find the angle between the plane:
x + y - 2z = 3 and 2x - 2y + z = 5
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between x + y - 2z = 3 and 2x - 2y + z = 5 is given by
$\cos\theta=\frac{(1)(2)+(1)(-2)+(-2)(1)}{\sqrt{1^2+1^2+(-2)^2}\sqrt{2^2+(-2)^2+1^2}}$
$=\frac{2-2-2}{\sqrt{1+1+4}\sqrt{4+4+1}}=\frac{-2}{\sqrt{6}\sqrt{9}}$
$=\frac{-2}{\sqrt{6}\sqrt{9}}=\frac{-2}{3\sqrt{6}}$
$\theta=\cos^{-1}\Big(\frac{-2}{3\sqrt{6}}\Big)$
View full question & answer→Question 93 Marks
Find the angle between the line $\frac{\text{x}+1}{2}=\frac{\text{y}}{3}=\frac{\text{z}-3}{6}$ and the plane 10x + 2y - 11z = 3.
AnswerThe given line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by.
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})\cdot(10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}||10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}|}$
$=\frac{20+6-66}{\sqrt{4+9+36}\sqrt{100+4+121}}$
$=\frac{-40}{(7)(15)}=\frac{-8}{21}$
$\theta=\sin^{-1}\Big(\frac{-8}{21}\Big)$
View full question & answer→Question 103 Marks
Find the coordinates of the point where the line through (3, -4, -5) and (2, -3, 1) corsses the plane 2x + y + z = 7.
AnswerThe equation of the through the points (3, -4, -5) and (2, -3, 1) is
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}$
$\Rightarrow\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}$
The coordinates of any point on this line are of the form
$\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}=\lambda$
$\Rightarrow\text{x}=-\lambda+3,\text{ y}=\lambda-4,\text{ z}=6\lambda-5$
$\Rightarrow5\lambda=10$
$\Rightarrow\lambda=2$
So, the coordinates of the point are
$(-\lambda+3,\lambda-4,6\lambda-5)$
$=(-2+3,2-4,6(2)-5)$
$=(1,-2,7)$
View full question & answer→Question 113 Marks
Find the distance of the point $(2, 3, 5)$ from the xy-plane.
AnswerWe know that, the distance (D) of a the point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by
$\text{D}=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ ...(\text{i})$
So, distance of point (2, 3, 5) from xy-plane (we know that equation of xy-plane is z = 0) is
$=\Bigg|\frac{(2)(0)+(3)(0)+(5)(1)+0}{\sqrt{(0)^2+(0)^2+(1)^2}}\Bigg|$ [Using (i)]
$=\frac{0+0+5}{\sqrt{0+0+1}}$
$=5\text{ unit}$
Distance of the point (2, 3, 5) from xy-plane = 5 unit
View full question & answer→Question 123 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x + 3y - z = 6
AnswerThe equation of the given plane is,2x + 3y - z = 6
Dividng both sides by 6, we get
$\frac{2\text{x}}{6}+\frac{3\text{y}}{6}-\frac{\text{z}}{6}=\frac{6}{6}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{2}+\frac{\text{z}}{-6}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=3;\text{ b}=2;\text{ c}=-6$
View full question & answer→Question 133 Marks
Find the angle between the line $\vec{\text{r}}=(2\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ and the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=5$
AnswerWe know that the angle $\theta$ between the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
Here,
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
So, $\sin\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}$
$=\frac{9}{\sqrt{29}\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{29}}$
$\theta=\sin^{-1}\Big(\frac{3\sqrt{3}}{\sqrt{29}}\Big)$
View full question & answer→Question 143 Marks
Find the angle between the plane:
2x + y - 2z = 5 and 3x - 6y - 2z = 7
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is given by
$\cos\theta=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{2^2+1^2+(-2)^2}\sqrt{3^2+(-6)^2+(-2)^2}}$
$=\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{4}{(3)(7)}=\frac{4}{21}$
$\theta=\cos^{-1}\Big(\frac{4}{21}\Big)$
View full question & answer→Question 153 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$2\text{x}-4\text{y}+3\text{z}=5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$
AnswerWe know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are 2x - 4y + 3z = 5 and $\text{x}+2\text{y}+\lambda\text{z}=5$
$\Rightarrow a_1 = 2; b_1 = -4; c_1 = 3; a_2 = 1; b_2 = 2$; $\text{c}_2=\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(2)(1)+(-4)(2)+(3)+(\lambda)=0$
$\Rightarrow2-8+3\lambda=0$
$\Rightarrow3\lambda=6$
$\Rightarrow\lambda=2$
View full question & answer→Question 163 Marks
Find the equation of the plane passing through the following points:
(2, 1, 0), (3, -2, -2) and (3, 1, 7)
AnswerThe equation of the plane passing through points (2, 1, 0), (3, -2, -2) and (3, 1, 7) is given by,$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$
$\Rightarrow-21(\text{x}-2)-9(\text{y}-1)+3\text{z}=0$
$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$
$\Rightarrow-21\text{x}-9\text{y}+3\text{z}+51=0$
$\Rightarrow21\text{x}+9\text{y}-3\text{z}=51$
$\Rightarrow7\text{x}+3\text{y}-\text{z}=17$
View full question & answer→Question 173 Marks
Find the vector and cartesian equations of a plane passing through the point (1, -1, 1) and normal to the line joining the points (1, 2, 5) and (-1, 3, 1)
AnswerSince the given plane passes through the point (1, -1, 1) and is normal to the line joining A(1, 2, 5) and B(-1, 3, 1)
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}})$
$=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
We know that the vector equation of the passing through a point $\overrightarrow{\text{a}}$ and normal to $\overrightarrow{\text{n}}$ is,
$\overrightarrow{\text{r}}\cdot\overrightarrow{\text{n}}=\overrightarrow{\text{a}}\cdot\overrightarrow{\text{n}}$
Substituting $\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{n}}=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ we get
$\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=-2-1-4$
$\Rightarrow\overrightarrow{\text{r}}\cdot\big[-(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})\big]=-7$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=7$
For cartesian form, we need to substitute $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation.
Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})=7$
$\Rightarrow2\text{x}-\text{y}+4\text{z}=7$
View full question & answer→Question 183 Marks
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.
AnswerLet $\alpha,\beta$ and $\gamma$ be the angles made by $\vec{\text{n}}$ with x, y and z-axes, respectively.
It is given that
$\alpha=\beta=\gamma$
$\Rightarrow\cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\text{l}=\text{m}=\text{n},$ where l, m, n are direction cosines of $\vec{\text{n}}$.
But $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{l}^2+\text{l}^2=1$
$\Rightarrow3\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}^2=\frac{1}{\sqrt{3}}$
So, $\text{l}=\text{m}=\text{n}=\frac{1}{\sqrt{3}}$
It is given that the length of the perpendicular of the plane from the origin, $\text{p}=5\sqrt{3}$
The normal form of the plane is $\text{lx}+\text{my}+\text{nz}=\text{p}$
$\Rightarrow\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=5\sqrt{3}$
$\Rightarrow\text{x}+\text{y}+\text{z}=5\sqrt{3}\ (\sqrt{3})$
$\Rightarrow\text{x}+\text{y}+\text{z}=15.$
View full question & answer→Question 193 Marks
Find the distance of the point P(-1, -5, -10) from the point of intersection of the line joining the points A(2, -1, 2) and B(5, 3, 4) with the plane x - y + z = 5.
AnswerEquation of the line through the points A(2, -1, 2) and B(5, 3, 4) is $\frac{\text{x}-2}{5-2}=\frac{\text{y}+1}{3+1}=\frac{\text{z}-2}{4-2}=\text{r}$
$\Rightarrow\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}=\text{r}$
$\Rightarrow\text{x}=3\text{r}+2,\text{ y}=4\text{r}-1,\text{ z}=2\text{r}+2$
Substituting these in the plane equation we get
$(3\text{r}+2)-(4\text{r}-1)+(2\text{r}+2)=5$
$\Rightarrow\text{r}=0$
$\Rightarrow\text{x}=2,\text{ y}=-1,\text{ z}=2$
Distance of (2, -1, 2) from (-1, -5, -10) is
$=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{3^2+4^2+12^2}$
$=\sqrt{169}=13$
View full question & answer→Question 203 Marks
Write the vector equation of the line passing through the point (1, -2, -3) and normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5.$
AnswerThe required line is normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5$ and it is parallel to the normal vector of the plane.
So, the required line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
It is given that the line passes through the point (1, -2, -3) whose position vector is given by $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
We know that the equation of the line passing through the point whose position vector is $\vec{\text{a}}$ and parrallel to the vector $\vec{\text{b}}$ is given by
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 213 Marks
Write the distance between the parallel planes $2x − y + 3z = 4$ and $2x − y + 3z = 18$.
AnswerThe given equation are
$2x − y + 3z = 4 ....(1)$
The second equation of the plane is
$2x − y + 3z = 18 .....(2)$
We know that 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 is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$
View full question & answer→Question 223 Marks
Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point $(\alpha,\beta,\gamma)$
AnswerLet a, b and c be the intercepts of the given plane on the coordinate axes.
Then the plane meets the coordinate axes at
A(a, 0, 0), B(0, b, c) and C(0, 0, c)
Given that the centroid of the triangle $=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\frac{\text{a}}{3}=\alpha,\frac{\text{b}}{3}=\beta,\frac{\text{c}}{3}=\gamma$
$\Rightarrow\text{a}=3\alpha,\text{b}=3\beta,\text{c}=3\gamma\ ...(\text{i})$
The equation of the plane whose intercepts on the coordinate axes are a, b and c are
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1$ [From (i)]
$\Rightarrow\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=3$
View full question & answer→Question 233 Marks
Find the angle between the plane:
2x - y + z = 4 and x + y + 2z = 3
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - y + z = 4 and x + y + 2z = 3 is given by
$\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+1^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}=\frac{3}{\sqrt{6}\sqrt{6}}$
$=\frac{3}{6}=\frac{1}{2}$
$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$
View full question & answer→Question 243 Marks
Find the distance of the point $2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}$ from the plane $\vec{\text{r}}\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9=0$
AnswerWe know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=-2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}; \vec{\text{n}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}};\text{d}=9$
So, the required distance, p
$=\frac{\big|(2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9\big|}{\big|3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}-9\big|}$
$=\frac{6+4-48-9}{\sqrt{9+16+144}}$
$=\frac{|-47|}{13}$
$=\frac{47}{13}\text{ units}$
View full question & answer→Question 253 Marks
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y - z = 1 and 3x - 4y + z = 5.
AnswerThe equation of any plane passing through the origin (0, 0, 0) is,
a(x - 0) + b(y - 0) + c(z - 0) = 0
ax + by + cz = 0 ...(i)
It is given that (i) is perpendicular to the planes x + 2y - z = 1 and 3x - 4y + z = 5. Then,
a + 2b - c = 0 ....(ii)
3a - 4b + c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}&\text{y}&\text{z}\\1&2&-1\\3&-4&1\end{vmatrix}=0$
⇒ -2x - 4y - 10z = 0
⇒ x + 2y + 5z = 0
View full question & answer→Question 263 Marks
Find the equation of a passing through the point (-1, -1, 2) and perpendicular to the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5.
Question 273 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=9$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{n}}_2=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6{\hat{\text{k}}\big|}\big|\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{2-16-12}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{-16}{(7)(3)}$
$=\frac{-16}{21}$
$\theta=\cos^{-1}\Big(\frac{-16}{21}\Big)$
View full question & answer→Question 283 Marks
Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x - 3y + 5z + 7 = 0. Also, find the distance between the two planes.
AnswerEquation of plane is parallel to 2x - 3y + 5z + 7 = 0 is of the form 2x - 3y + 5z = d
Above plane is passing through (3, 4, -1)
So, substitute above point in the equation, we get
6 - 12 - 5 = d
d = -11
So, palne equation is 2x - 3y + 5z = -11
Distance between planes is given by
$\Big|\frac{-7+11}{\sqrt{4+9+25}}\Big|=\frac{4}{\sqrt{38}}$
View full question & answer→Question 293 Marks
Write the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ in normal form.
AnswerThe given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ or $\vec{\text{r}}.\vec{\text{n}}=14$, where $\vec{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+9+36}=7$
For reducing the given equation to normal form, we need to divide it by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{14}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}{7}\Big)=\frac{14}{7}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2,$ which is the required normal form.
View full question & answer→Question 303 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})=1$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}})=4$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{n}}_2=-\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})\cdot(-\hat{\text{i}}+\vec{\text{j}}+0\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big|}$
$=\frac{-2-3}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}$
$=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$
View full question & answer→Question 313 Marks
A plabne makes intercepts -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.
AnswerWe know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
So, the equation of the plane which makes intercepts -6, 3, 4 on the x-axis, y-axis and z-axis, respecticely is,
$\frac{\text{x}}{-6}+\frac{\text{y}}{3}+\frac{\text{z}}{4}=1$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}=12$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}+12=0$
$\therefore$ Length of the perpendicular from (0, 0, 0) to the plane 2x - 4y - 3z + 12 = 0
$=\Bigg|\frac{2\times0-4\times0-3\times0+12}{\sqrt{12^2+(-4)^2+(-3)^2}}\Bigg|$
$=\bigg|\frac{12}{\sqrt{4+16+9}}\bigg|$
$=\frac{12}{\sqrt{29}}\text{ units}$
Thus, the length of the perpendicular from the origin to the plane is $\frac{12}{\sqrt{29}}\text{ units}.$
View full question & answer→Question 323 Marks
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$
AnswerThe equation of the family of plane parallel to $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$ is,
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{d}\ ...(\text{i})$
If it passes through (a, b, c) then
$(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=\text{d}$
$=\text{a}+\text{b}+\text{c}$
Substituting a + b + c = d in (i) we get
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
$\text{x}+\text{y}+\text{z}=\text{a}+\text{b}+\text{c}$ as the equation of the required plane.
View full question & answer→Question 333 Marks
Find a unit normal vector to the plane x + 2y + 3z - 6 = 0
AnswerThe given equation of the plane is
x + 2y + 3z - 6 = 0
x + 2y + 3z = 6
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=6$
Or $\vec{\text{r}}\cdot\vec{\text{n}}=6$
When, $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ ...(\text{i})$
Now, $|\vec{\text{n}}|=\sqrt{1^2+2^2+3^2}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
Unit vector to the plane, $\hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{14}}$
$=\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}$
View full question & answer→Question 343 Marks
Find the vector equation of the plane with intercepts 3, -4 and 2 on x, y and z-axis respectively.
AnswerThe equation of the plane in the intercept form is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$ where a, b and c are the intercepts on the x, y and z-axis, respectively.
It is given that intercepts made by the plane on the X, Y and Z-axis are 3, -4 and 2, respectively.
$\therefore$ a = 3, b = -4, c = 2
Thus, the equation of the plane is
$\frac{\text{x}}{3}+\frac{\text{y}}{(-4)}+\frac{\text{z}}{2}=1$
$\Rightarrow4\text{x}-3\text{y}+6\text{z}=12$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(4\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}})=12$
This is the vector form of the equation of the given plane.
View full question & answer→Question 353 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x - y + z = 5
AnswerEquation of the given plane is,2x - y + z = 5
Dividng both sides by 5, we get
$\frac{2\text{x}}{5}+\frac{-\text{y}}{5}+\frac{\text{z}}{5}=\frac{5}{5}$
$\Rightarrow\frac{\text{x}}{\big(\frac{5}{2}\big)}+\frac{\text{y}}{-5}+\frac{\text{z}}{5}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=\frac{5}{2};\text{ b}=-5;\text{ c}=5$
View full question & answer→Question 363 Marks
Find the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ to the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}.$ Hence, of otherwise, deduce the length of the perpendicular.
AnswerLet M be the foot of the perpendicular of the point $P(5, 4, 2)$ on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}$
Therefore, its equation is
$\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}=\text{r}$
Then, M is in the form $(2r - 1, 3r + 3, -r + 1)$
Direction ratios of MP are $2r - 1 - 5, 3r + 3 - 4, -r + 1 - 2$ or $2r - 6, 3r - 1, -r - 1$.
Since MP is perpendicular to the given line $(2, 3, -1), 2 (2r - 6) + 3(3r - 1) -1(-r - 1) = 0$ $($Because $a_1, a_2, + b_1, b_2, + c_1, c_2, = 0)$
$\Rightarrow 4r - 12 + 9r - 3 + 1 + 1 = 0$
$\Rightarrow 14r - 14 = 0$
$\Rightarrow r = 1$
So, $M = (2r - 1, 3r + 3, -r + 1) = (2 (1) - 1, 3(1) + 3, -1 + 1) = (1, 6, 0)$
Length of the perperndicular,
$\text{MP}=\sqrt{(1-5)^2+(6-4)^2+(0-2)^2}\\=\sqrt{16+4+4}=\sqrt{24}=2\sqrt{6}\text{ units}$
View full question & answer→Question 373 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=7$ and $\vec{\text{r}}\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=26$
AnswerWe know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}, \vec{\text{n}}_2=\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}}$
The given planes are perpendicular.
$\Rightarrow\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
$\Rightarrow(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=0$
$\Rightarrow\lambda+4-21=0$
$\Rightarrow\lambda-17=0$
$\Rightarrow\lambda=17$
View full question & answer→Question 383 Marks
Reduce the equation 2x - 3y - 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.
AnswerGiven equation of plane is,
2x - 3y - 6z = 14
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})=14$
Dividing the equation by $\sqrt{(2)^2+(-3)^2+(-6)^2}$
$\vec{\text{r}}\cdot\frac{(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})}{\sqrt{4+9+36}}=\frac{14}{\sqrt{4+9+36}}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=\frac{14}{7}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2\ ...(\text{i})$
We know that the vector equation of a plane with distance d from origin and normal to unit vector $\hat{\text{n}}$ is given by
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{d}}\ ...(\text{ii})$
Comparing (i) and (ii),
d = 2 and
$\hat{\text{n}}=\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}$
So, distance of plane from origin = 2 unit
Direction cosine of normal to plane $=\frac{2}{7},-\frac{3}{7},\frac{6}{7}$
View full question & answer→Question 393 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
zx-plane
AnswerDirection ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the zx-plane y = 0
$\Rightarrow-3\text{r}+1=0$
$\Rightarrow\text{r}=\frac{1}{3}$
$\Rightarrow\text{x}=2\Big(\frac{1}{3}\Big)+5=\frac{17}{3}$
$\Rightarrow\text{z}=5\Big(\frac{1}{3}\Big)+6=\frac{23}{3}$
Hence, the corrdinates of the point are $\Big(\frac{17}{3},0,\frac{23}{3}\Big)$
View full question & answer→Question 403 Marks
Write the ratio in which the plane $4x + 5y − 3z = 8$ divides the line segment joining the points $(−2, 1, 5)$ and $(3, 3, 2)$.
AnswerWe know that the ratio in which the plane $ax + by + cz + d = 0$ divides the line sebment joining
$(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{-(\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d})}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
Here, $a = 4,b = 5,c = -3,d = -8,x_1 = -2,y_1 = 1,z_1 = 5,x_2 = 3,y_2 = 3,z_2 = 2$
So, the required ratio
$=\frac{-(4(-2)+5(1)-3(5)-8)}{4(3)+5(3)-3(2)-8}$
$=\frac{-(-8+5-15-8)}{12+15-6-8}$
$=\frac{26}{13}$
$=\frac{2}{1}$ or $2 :1.$
View full question & answer→Question 413 Marks
Find the direction cosines of the unit vector perpendicular to the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$ passing through the origin.
AnswerGiven equation of the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$
Thus, the direction ratios normal to the plane are 6, -3 and -2
Hence, the direction cosines to the normal to the plane are
$=\frac{6}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}}$
$=\frac{6}{7},\frac{-3}{7},\frac{-2}{7}$
$=\frac{-6}{7},\frac{3}{7},\frac{2}{7}$
The direction cosines of the unit vector perpendicular to the plane are same as the direction cosines of the normal to the plane.
Thus, the direction cosined of the unit vector perpendicular to the plane are: $\frac{-6}{7},\frac{3}{7},\frac{2}{7}$
View full question & answer→Question 423 Marks
Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.
AnswerThe given equation of the plane is
2x − 3y + 4z = 12
Dividing both sides by 12, we get
$\Rightarrow\frac{2\text{x}}{\text{12}}+\frac{-3\text{y}}{\text{12}}+\frac{4\text{z}}{\text{12}}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{\text{6}}+\frac{\text{y}}{-\text{4}}+\frac{\text{z}}{\text{3}}=1\ ....(1)$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(2)$
Comparing (1) and (2), we get
a = 6, b = -4 and c = 3.
View full question & answer→Question 433 Marks
Show that the following planes are at right angles.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=3$
AnswerWe know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{n}}_2=-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\vec{\text{n}}_1\cdot\vec{\text{n}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=-2+1+1=0$
So, the given planes are perpendicular.
View full question & answer→Question 443 Marks
Find the equation of the plane passing throught the point (2, 4, 6) and making equal intercepts on the coordinate axes.
AnswerIntercepts on the coordinate axes are equal.
We know that, if a, b, c are Intercepts on coordinate axes by a plane, then equationb of the plane is given by,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
Here, it is given that a = b = c = p (say)
$\frac{\text{x}}{\text{p}}+\frac{\text{y}}{\text{p}}+\frac{\text{z}}{\text{p}}=1$
$\frac{\text{x}+\text{y}+\text{z}}{\text{p}}=1$
$\text{x}+\text{y}+\text{z}=\text{p}\ ...(\text{i})$
It is given that plane is passing through the point (2, 4, 6), so using equation (i)
x + y + z = p
2 + 4 + 6 = p
12 = p
Put, value of p in equation (i)
x + y + z = 12
So, the required equation of the plane is given by,
x + y + z = 12
View full question & answer→Question 453 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$
Answer$\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$
Plane is passing through $(\hat{\text{i}}-\hat{\text{j}})$ and parallel to
$\text{b}(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$ and $\text{c}(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\4&-2&3\end{vmatrix}$
$\text{n}=5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{r}}\cdot\text{n}=(\hat{\text{i}}-\hat{\text{j}})\cdot(5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}})$
$=5-1=4$
$\text{r}\cdot(5\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}})=4$
View full question & answer→Question 463 Marks
Write the normal form of the equation of the plane 2x - 3y + 6z + 14 = 0
AnswerThe given equation of the plane is,
2x - 3y + 6z + 14 = 0
2x - 3y + 6z + 14 = 0 ...(i)
Now, $\sqrt{2^2+(-3)^2+(6)^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Dividing (i) by 7, we get
$\frac{2}{7}\text{x}-\frac{3}{7}\text{y}+\frac{6}{7}\text{z}=2$
This is the normal form of the given equation of the plane.
View full question & answer→Question 473 Marks
Find the equation of rthe planes parallel to the plane x + 2y - 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).
AnswerThe equation of the plane parallel to the given plane is
x + 2y - 2z + k = 0 ....(i)
It is given that plane (i) is at a distance of 2 unit from (2, 1, 1).
$\Rightarrow\frac{|2+2-2+\text{k}|}{\sqrt{1^2+2^2+(-2)^2}}=2$
$\Rightarrow\frac{|2+\text{k}|}{3}=2$
$\Rightarrow|2+\text{k}|=6$
$\Rightarrow2+\text{k}=6,{ 2}+\text{k}=-6$
$\Rightarrow\text{k}=4,\text{k}=-8$
Substi9tuting these two values one by one in (i) we get
x + 2y - 2z + 4 = 0 and x + 2y - 2z - 8 = 0, which are the equatioms of the required planes.
View full question & answer→Question 483 Marks
Find the angle between the plane:
x - y + z = 5 and x + 2y + z = 9
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between x - y + z = 5 and x + 2y + z = 9 is given by
$\cos\theta=\frac{(1)(1)+(-1)(2)+(1)(1)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{1^2+2^2+1^2}}$
$=\frac{1-2+1}{\sqrt{1+1+1}\sqrt{1+4+1}}$
$=\frac{0}{\sqrt{3}\sqrt{6}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$
View full question & answer→Question 493 Marks
Find the angle between the plane:
2x - 3y + 4z = 1 and -x + y = 4
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - 3y + 4z = 1 and -x + y + 0z = 4 is given by
$\cos\theta=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^2+(-3)^2+4^2}\sqrt{(-1)^2+1^2+0^2}}$
$=\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$
View full question & answer→Question 503 Marks
The line $\vec{\text{r}}=\hat{\text{i}}+\lambda(2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}})$ is parallel to the plane $\vec{\text{r}}\cdot(\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})=4.$ Find m.
AnswerThe given line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
If the line is parallel to the plane, the normal to the plane is perpendicular to the line.
$\Rightarrow\vec{\text{b}}\perp\vec{\text{n}}$
$\Rightarrow\vec{\text{b}}\cdot\vec{\text{n}}=0$
$\Rightarrow(2\hat{\text{i}}-\text{m}\hat{\text{j}}-3\hat{\text{k}})\cdot(\text{m}\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})=0$
$\Rightarrow2\text{m}-3\text{m}-3=0$
$\Rightarrow-\text{m}-3=0$
$\Rightarrow\text{m}=-3$
View full question & answer→Question 513 Marks
Write the equation of the plane whose intercepts on the coordinate axes are 2, -3 and 4
AnswerGiven, intercepts on the coordinate axes are 2, -3 and 4
We know that,
The equation of a plane whose intercept onb the coordinate axes are a, b and c
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{i})$
Here, a = 2, b = -3, c = 4
So,
Equation of required plane is,
$\frac{\text{x}}{2}+\frac{\text{y}}{-3}+\frac{\text{z}}{4}=1$
$\frac{6\text{x}-4\text{y}+3\text{z}}{12}=1$
${6\text{x}-4\text{y}+3\text{z}=}{12}$
View full question & answer→Question 523 Marks
A plane meets the coordinate axes at A, B and C, respectively, such that the centriod of triangle ABC is (1, -2, 3). Find the equation of the plane.
AnswerLet a, b and c be the intercepts of the given plane on the coordinate axes.
Then the plane meets the coordinate axes at
A(a, 0, 0), B(0, b, 0) and C(0, 0, c)
Given that the centroid of the triangle is (1, -2, 3)
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\frac{\text{a}}{3}=1,\frac{\text{b}}{3}=-2,\frac{\text{c}}{3}=3$
$\Rightarrow\text{a}=3,\text{b}=-6,\text{c}=9\ ...(\text{i})$
Equation of the plane whose intercepts on the coordinate axes a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{-6}+\frac{\text{z}}{9}=1$ [From (i)]
$\Rightarrow6\text{x}-3\text{y}+2\text{x}=18$
View full question & answer→Question 533 Marks
Show that the normals to the following parirs of planes are perpendicular other.
x - y + z - 2 = 0 and 3x + 2y - z + 4 = 0
AnswerLet $\vec{\text{n}_1}$ and $\vec{\text{n}_2}$ be the vectors which are normals to the planes x - y + z = 2 and 3x + 2y - z = -4 respectively.
The given equations of the planes are
x + y + z = 2
3x + 2y - z = -4
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=8,$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow\vec{\text{n}_1}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\vec{\text{n}_2}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $\vec{\text{n}_1}\cdot\vec{\text{n}_2}=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=3-2-1=0$
So, the normals to the given planes are perpendicular to each other.
View full question & answer→Question 543 Marks
Show that the following point are coplanar:
(0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1)
AnswerThe equation of the plane passing through points (0, 4, 3), (-1, -5, -3), (-2, -2, 1) is, $\begin{vmatrix}\text{x}-0&\text{y}-4&\text{z}-3\\-1-0&-5-4&-3-3\\-2-0&-2-4&1-3\end{vmatrix}=0$ $\Rightarrow\begin{vmatrix}\text{x}&\text{y}-4&\text{z}-3\\-1&-9&-6\\-2&-6&-2\end{vmatrix}=0$ $\Rightarrow-18\text{x}+10(\text{y}-4)-12(\text{z}-3)=0$ $\Rightarrow9\text{x}-5(\text{y}-4)+6(\text{z}-3)=0$ $\Rightarrow9\text{x}-5\text{y}+\text{z}+2=0$ Substituting the last points (1, 1, -1) (it means x = 1; y = 1; z = -1) in this plane equation, we get9(1) - 5(1) + 6(-1) + 2 = 0
⇒ 4 - 4 = 0 ⇒ 0 = 0 So, the plane equation is satisfied by the points (1, 1, -1) So, the given pointsa are coplanar.
View full question & answer→Question 553 Marks
Obtain the equation of the plane passing through the point (1, - 3, -2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
AnswerThe equation of any plane passing through (1, -3, -2) is,
a(x - 1) + b(y + 3) + c(z + 2) = 0 ....(i)
It is given that (i) is perpendicular to each of the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8. Then,
a + 2b + 2c = 0 ....(ii)
3a + 3b + 2c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-1&\text{y}+3&\text{z}+2\\1&2&2\\3&3&2\end{vmatrix}=0$
⇒ -2(x - 1) + 4(y + 3) - 3(z + 2) = 0
⇒ -2x + 2 + 4y + 12 - 3z - 6 = 0
⇒ 2x - 4y + 3z - 8 = 0
View full question & answer→Question 563 Marks
Find the angle between the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{-1}=\frac{\text{z}-3}{2}$ and the plane 3x + 4y + z + 5 = 0
AnswerThe given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and the given plane id normal to the vector $\vec{\text{n}}=3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}|\vec{|\text{n}}|}$
$=\frac{(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}})}{|3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}||3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}=\frac{7}{\sqrt{14}\sqrt{26}}$
$=\frac{7}{\sqrt{2}\sqrt{7}\sqrt{2}\sqrt{13}}=\frac{\sqrt{7}}{\sqrt{52}}=\sqrt{\frac{7}{52}}$
$\theta=\sin^{-1}\Big(\sqrt{\frac{7}{52}}\Big)$
View full question & answer→Question 573 Marks
Find the equation of the plane passing through the following point:
(-5, 0, -6), (-3, 10, -9) and (-2, 6, -6)
AnswerThe equation of the plane passing through points (-5, 0, -6), (-3, 10, -9) and (-2, 6, -6) is given by,
$\begin{vmatrix}\text{x}+5&\text{y}-0&\text{z}+6\\-3+5&10-0&-9+6\\-2+5&6-0&-6+6\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}+5&\text{y}&\text{z}+6\\2&10&-3\\3&6&0\end{vmatrix}=0$
$\Rightarrow18(\text{x}+5)-\text{y}-18(\text{z}+6)=0$
$\Rightarrow2(\text{x}+5)-\text{y}-2(\text{z}+6)=0$
$\Rightarrow2\text{x}+10-\text{y}-2\text{z}-12=0$
$\Rightarrow2\text{x}-\text{y}-2\text{z}-2=0$
View full question & answer→Question 583 Marks
Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.
AnswerThe equation of the plane parallel to the plane ZOX is,
y = b ....(i), where b is a constant.
It is given that this plane passes through (0, 3, 0). So,
3 = b
Substituting this value in (i), we get
y = 3, which is the required equation of the plane.
View full question & answer→Question 593 Marks
Show that the following point are coplanar:
(0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0)
AnswerThe equation of the plane passing through points (0, -1, 0), (2, 1, -1) and (1, 1, 1) is given by, $\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\2-0&1+1&-1-0\\1-0&1+1&1-0\end{vmatrix}=0$ $\Rightarrow\begin{vmatrix}\text{x}&\text{y}+1&\text{z}\\2&2&-1\\1&2&1\end{vmatrix}=0$ $\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$ $\Rightarrow4\text{x}-3\text{y}+2\text{z}-3=0$ Substituting the last points (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get4(3) - 3(3) + 2(0) - 3 = 0
⇒ 12 - 12 = 0 ⇒ 0 = 0 So, the plane equation is satisfied by the points (3, 3, 0) So, the given pointsa are coplanar.
View full question & answer→Question 603 Marks
Show that the following planes are at right angles.
$x - 2y + 4z = 10$ and $18x + 17y + 4z = 49$
AnswerWe know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are $x - 2y + 4z = 10$ and $18x + 17y + 4z = 49$
$\Rightarrow a_1 = 1; b_1 = -2; c_1 = 4; a_2 = 18; b_2 = 17; c_2 = 4$
Now,
$a_1a_2 + b_1b_2 + c_1c_2$
$= (1)(18) + (-2)(17) + (4)(4)$
$= 18 - 34 + 16$
$= 0$
So, the given planes are perpendicular.
View full question & answer→Question 613 Marks
Show that the normals to the following parirs of planes are perpendicular other.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$
AnswerLet $\vec{\text{n}_1}$ and $\vec{\text{n}_2}$ be the vectors which are normals to the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$ respectively.The given equations of the planes are
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$
$\Rightarrow\vec{\text{n}_1}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}),$
$\Rightarrow\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$
Now, $\vec{\text{n}_1}\cdot\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$
$=4+2-6=0$
So, the normals to the given planes are perpendicular to each other.
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