Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.5

Last Updated : 11 Feb, 2021

Question 1. Find the vector equation of the plane passing through the points (1, 1, 1), (1, -1, 1) and (-7, -3, -5)

Solution:

Given that, plane is passing through
(1, 1, 1), (1, -1, 1) and (-7, -3, -5)
We know that, equation of plane passing through 3 points,
$\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$
$\begin{vmatrix}x-1 & y-1 & z-1\\ 1-1 & -1-1 & 1-1\\ -7-1 & -3-1 & -5-1\end{vmatrix}=0$
$\begin{vmatrix}x-1 & y-1 & z-1\\ 0 & -2 & 0\\ -8 & -4 & -6\end{vmatrix}=0$
(x – 1)(12 – 0) – (y – 1)(0 – 0) + (z – 1)(0 – 16) = 0
(x – 1)(12) – (y – 1)(0) + (z – 1)(-16) = 0
12x – 12 – 0 – 16z + 16 = 0
12x – 16z + 4 = 0
Dividing by 4,
3x – 4z + 1 = 0
$(x\hat{i}+y\hat{j}+z\hat{k})(3\hat{i}+0\hat{j}-4\hat{k})+1=0\\ \vec{r}.(3\hat{i}-4\hat{k})+1=0$
Equation of the required plane,
$\vec{r}.(3\hat{i}-4\hat{k})+1=0$

Question 2. Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).

Solution:

Let P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3) be the three points on a plane having position vectors $\vec{p},\vec{q}\ and\ \vec{s}$ respectively. Then the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ are in the same plane. Therefore, $\overrightarrow{PQ}\times\overrightarrow{PR}$ is a vector perpendicular to the plane.
Let = $\vec{n} = \overrightarrow{PQ}\times\overrightarrow{PR}$
$\overrightarrow{PQ}=(-2-2)\hat{i}+(-3-5)\hat{j}+(5-(-3))\hat{k}\\ \overrightarrow{PQ}=-4\hat{i}-8\hat{j}+8\hat{k}$
Similarly,
$\overrightarrow{PR}=(5-2)\hat{i}+(3-5)\hat{j}+(-3-(-3))\hat{k}\\ \overrightarrow{PQ}=3\hat{i}-2\hat{j}+0\hat{k}$
Thus
$\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}\\ =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ -4 & -8 & 8\\ 3 & -2 & 0\end{vmatrix}\\ =16\hat{i}+24\hat{j}+32\hat{k}$
The plane passes through the point P with position vector $\vec{p}=2\hat{i}+5\hat{j}-3\hat{k}$
Thus, its vector equation is
$(\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\$
$(\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-(32+120-96)=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-56=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})=56\\ ⇒\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7$

Question 3. Find the vector equation of the plane passing through the points A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from origin, prove that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

Solution:

Let A(a, 0, 0), B(0, b, 0) and C(0, 0, c) be three points on a plane having their position vectors $\vec{a},\vec{b}\ and\ \vec{c}$ respectively. Then vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are in the same plane. Therefore, $\overrightarrow{AB}\times\overrightarrow{AC}$ is a vector perpendicular to the plane.
Let $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$
$\overrightarrow{AB}=(0-a)\hat{i}+(b-0)\hat{j}+(0-0)\hat{k}\\ \overrightarrow{AB}=-a\hat{i}+b\hat{j}+0\hat{k}$
Similarly,
$\overrightarrow{AC}=(0-a)\hat{i}+(0-0)\hat{j}+(c-0)\hat{k}\\ \overrightarrow{AC}=-a\hat{i}+0\hat{j}+c\hat{k}$
Thus
$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$
$\hat{i}\ \ \ \hat{j}\ \ \ \hat{k}$
= | -a b 0 |
-a 0 c
$\vec{n}=bc\hat{i}+ac\hat{j}+ab\hat{k}$
$⇒\hat{n}=\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}$
The plane passes through the point P with position vector $\vec{a}=a\hat{i}+0\hat{j}+0\hat{k}$
Thus, the vector equation in the normal form is
$\{\vec{r}-\left(a\hat{i}+0\hat{j}+0\hat{k}\right)\}.\left(\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\right)=0\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}...(1)$
The vector equation of a plane normal to the unit vector $\hat{n}$ and at a distance ‘d’ from the origin is $\vec{r}.\hat{n}=d$ ….(2).
Given that the plane is at a distance ‘p’ from the origin.
Comparing equations (1) and (2), we have,
d = p = $\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\\ ⇒\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

Question 4. Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).

Solution:

Let P(1, 1, -1), Q(6, 4, -5) and R(-4, -2, 3) be three points on a plane having position vectors $\vec{p},\ \vec{q}\ and\ \vec{s}$ respectively. Then the vectors $\overrightarrow{PQ}\ and\ \overrightarrow{PR}$ are in the same plane. Therefore, $\overrightarrow{PQ}\times\overrightarrow{PR}$ is a vector perpendicular to the plane.
Let $\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$
$\overrightarrow{PQ}=(6-1)\hat{i}+(4-1)\hat{j}+(-5-(-1))\hat{k}\\ \overrightarrow{PQ}=5\hat{i}+3\hat{j}-4\hat{k}$
Similarly,
$\overrightarrow{PR}=(-4-1)\hat{i}+(-2-1)\hat{j}+(3-(-1))\hat{k}\\ \overrightarrow{PR}=-5\hat{i}-3\hat{j}+4\hat{k}$
Thus
Here, $\overrightarrow{PQ}=-\overrightarrow{PR}$
Therefore, the given points are collinear.
Thus, $\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}$ where, 5a + 3b – 4c = 0
The plane passes through the point P with position vector $\vec{p}=\hat{i}+\hat{j}-\hat{k}$
Thus, its vector equation is
$\{\vec{r}-(\hat{i}+\hat{j}-\hat{k})\}.(a\hat{i}+b\hat{j}+c\hat{k})=0$ , where, 5a + 3b – 4c = 0

Question 5. Find the vector equation of the plane passing through the points

$3\hat{i}+4\hat{j}+2\hat{k},\ 2\hat{i}-2\hat{j}-\hat{k}\ \ and\ \ 7\hat{i}+6\hat{k}$

Solution:

Let A, B, C be the points with position vector $(3\hat{i}+4\hat{j}+2\hat{k}),(2\hat{i}-2\hat{j}-\hat{k})\ and\ (7\hat{i}+6\hat{k})$
respectively. Then
$\overrightarrow{AB}$ = Position vector of B – Position vector of A
$=(2\hat{i}-2\hat{j}-\hat{k})-(3\hat{i}+4\hat{j}+2\hat{k})\\ =2\hat{i}-2\hat{j}-\hat{k}-3\hat{i}-4\hat{j}-2\hat{k}\\ =-\hat{i}-6\hat{j}-3\hat{k}$
$\overrightarrow{BC}$ = Position vector of C – Position vector of B
$=(7\hat{i}+6\hat{k})-(2\hat{i}-2\hat{j}-\hat{k})\\ =7\hat{i}+6\hat{k}-2\hat{i}+2\hat{j}+\hat{k}\\ =5\hat{i}+2\hat{j}+7\hat{k}$
A vector normal to A, B, C is a vector perpendicular to $\overrightarrow{AB}\times\overrightarrow{BC}$
$\vec{n}=\overrightarrow{AB}\times\overrightarrow{BC}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -1&-6&-3\\ 5&2&7\end{vmatrix}\\ \vec{n}=\hat{i}(-42+6)-\hat{j}(-7+15)+\hat{k}(-2+30)\\ =-36\hat{i}-8\hat{j}+28\hat{k}$
As we know that, equation of a plane passing through vector $\vec{a}$ and perpendicular to vector $\vec{n}$ is given by,
$\vec{r}.\vec{n}=\vec{a}.\vec{n}\ \ \ ...(1)$
Put $\vec{a}$ and $\vec{n}$ in equation (1)
$\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k})=(3\hat{i}+4\hat{j}+2\hat{k})(-36\hat{i}-8\hat{j}+28\hat{k})$
= (3)(-36) + (4)(-8) + (2)(28)
= -108 – 32 + 56
= -140 + 56
$\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k})$ = -84
Dividing by (-4), we will get
$\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21$
Equation of required plane is,
$\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21$