RD Sharma Class 12 Exercise 28.6 The Plane Solutions Maths

RD Sharma Class 12 Exercise 28.6 The Plane Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 25, 2022 11:42 AM IST

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RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise
The Plane Excercise: 28.6
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma class 12th exercise 28.6 contains solutions prepared by a group of subject experts who have years of experience in the field. Moreover, every answer goes through quality checks to ensure that the solutions provided provide the best information.

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Excercise: 28.6

The Plane exercise 28.6 question 1 (i)

Answer:
$\theta=\cos ^{-1} \frac{-5}{\sqrt{58}}$
Hint:
$\cos \theta=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}$
Given:
$\overrightarrow{\boldsymbol{r}} \cdot(\mathbf{2} \hat{\boldsymbol{\imath}}-3 \hat{\boldsymbol{\jmath}}+\mathbf{4} \widehat{\boldsymbol{k}})=\mathbf{1} \text { And } \vec{r} \cdot(-\hat{\imath}+\hat{\jmath})=4$
Solution:
$\begin{aligned} &\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \\ &\vec{n}_{1}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} \\ &\vec{r} \cdot(-\hat{\imath}+\hat{\jmath})=4 \\ &\vec{n}_{2}=-\hat{\imath}+\hat{\jmath} \\ &\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=\frac{(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}) \cdot(-\hat{\imath}+\hat{\jmath})}{\sqrt{2^{2}+(-3)^{2}+(4)^{2}} \cdot \sqrt{(-1)^{2}+\left ( 1 \right )^{2}}} \\ &=\frac{-2-3}{\sqrt{29 \times \sqrt{2}}}=\frac{-5}{\sqrt{58}} \\ \end{aligned}$
$\theta=\cos ^{-1}\frac{-5}{\sqrt{58}}$

The Plane exercise 28.6 question 1 (ii)

Answer:
$\cos ^{-1}\left ( -\frac{4}{21} \right )$
Hint:
$\cos \theta=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}$
Given:
$\overrightarrow{\boldsymbol{r}} \cdot(\mathbf{2} \hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}+2 \widehat{\boldsymbol{k}})=\mathbf{6} \text { And } \vec{r} \cdot(3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k})=9$
Solution:
$\begin{aligned} &\vec{n}_{1}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ &\vec{n}_{2}=3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k} \\ &\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=\frac{(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \cdot(3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k})}{|2 \hat{\imath}-\hat{\jmath}+2 \hat{k}| \cdot|3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k}|} \\ &=\frac{6-6-4}{\sqrt{2^{2}+(-1)^{2}+2^{2}} \cdot \sqrt{3^{2}+6^{2}+(-2)^{2}}}=\frac{-4}{3 \times 7}=-\frac{4}{21} \\ &\theta=\cos ^{-1}\left(-\frac{4}{21}\right) \end{aligned}$

The Plane exercise 28.6 question 1 (iii)

Answer:
$\cos ^{-1}\left ( -\frac{16}{21} \right )$
Hint:
$\cos \theta=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}$
Given:
$\vec{r} \cdot(2 \hat{\boldsymbol{\imath}}+3 \hat{\boldsymbol{\jmath}}-6 \hat{\boldsymbol{k}})=\mathbf{5} \text { And } \vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})=9$
Solution:
$\begin{aligned} &\vec{n}_{1}=2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k} \\ &\vec{n}_{2}=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \\ &\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=\frac{(2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k}) \cdot(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})}{|2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k}| \cdot|\hat{\imath}-2 \hat{\jmath}+2 \hat{k}|} \\ &=\frac{2-6-12}{\sqrt{4+9+36} \cdot \sqrt{1+4+4}}=\frac{-16}{3 \times 7} \\ &=-\frac{16}{21} \\ &\theta=\cos ^{-1}\left(-\frac{16}{21}\right) \end{aligned}$

The Plane exercise 28.6 question 2 (i)

Answer:
$\frac{\pi }{3}$
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:
$2 x-y+z=4 \text { and } x+y+2 z=3$
Solution:
$\begin{aligned} &2 x-y+z=4 \\ &a_{1}=2, b_{1}=-1, c_{1}=1 \\ &x+y+2 z=3 \\ &a_{2}=1, b_{2}=1, c_{2}=2 \\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{2(1)+(-1)(1)+(1)(2)}{\sqrt{4+1+1} \cdot \sqrt{1+1+4}} \\ &=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{1}{2} \\ & \end{aligned}$
$\theta=\frac{\pi}{3}$

The Plane exercise 28.6 question 2 (ii)

Answer:
$\cos ^{-1}\left ( -\frac{\sqrt{2}}{3\sqrt{3}} \right )$
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1} 2} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2} 2}}$
Given:
$x+y-2z=3 And 2x-2y+z=5$
Solution:
$\begin{aligned} &x+y-2 z=3 \\ &a_{1}=1, b_{1}=1, c_{1}=-2 \end{aligned}$
$\qquad \begin{array}{c} 2 x-2 y+z=5 \\ a_{2}=2, b_{2}=-2, c_{2}=1 \\ \cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ = \frac{(1)(2)+(1)(-2)+(-2)(1)}{\sqrt{(1)^{2}+(1)^{2}+(-2)^{2}} \cdot \sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}} \\ = \frac{-2}{\sqrt{6} \times 3} \\ = \frac{-\sqrt{2}}{3 \sqrt{3}} \\ \theta=\cos ^{-1}\left(-\frac{\sqrt{2}}{3 \sqrt{3}}\right) \end{array}$

The Plane exercise 28.6 question 2 (iii)

Answer:
$\frac{\pi }{2}$
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:
$x-y+z=5\hspace{0.2cm} And\hspace{0.2cm} x+2y+z=9$
Solution:
Here,
$\begin{aligned} &a_{1}=1, b_{1}=-1, c_{1}=1, \text { comparing with } x-y+z=5 \\ &a_{2}=1, b_{2}=2, c_{2}=1, \text { comparing with } x+2 y+z=9 \\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{(1)(1)+(-1)(2)+(1)(1)}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{(1)^{2}+(2)^{2}+(1)^{2}}} \\ &=0 \\ &\theta=\frac{\pi}{2} \end{aligned}$

The Plane exercise 28.6 question 2 (iv)

Answer:
$\cos ^{-1}\left ( -\frac{\sqrt{5}}{\sqrt{58}} \right )$
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1} 2}^{2} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2} 2}}$
Given:
$2x-3y+4z=1 \hspace{0.2cm}And \hspace{0.2cm} -x+y=4$
Solution:
Here,
$\begin{aligned} &a_{1}=2, b_{1}=-3, c_{1}=4, \text { on comparing with } 2 x-3 y+4 z=1 \\ &a_{2}=-1, b_{2}=1, c_{2}=0, \text { on comparing with }-x+y=4 \\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^{2}+(-3)^{2}+\left(4^{2}\right)} \cdot \sqrt{(-1)^{2}+(1)^{2}}} \\ &=\frac{-5}{\sqrt{29} \times \sqrt{2}} \\ &=-\frac{\sqrt{5}}{\sqrt{58}} \\ & \end{aligned}$
$\theta=\cos ^{-1}\left(-\frac{\sqrt{5}}{\sqrt{58}}\right)$

The Plane exercise 28.6 question 2 (v)

Answer:
$\cos ^{-1}\left ( \frac{4}{21} \right )$
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:
$2x+y+2z=5 And 3x-6y-2z$
Solution:
Here,
$\begin{aligned} &a_{1}=2, b_{1}=1, c_{1}=-2, \text { on comparing with } 2 x+y-2 z=5 \\ &a_{2}=3, b_{2}=-6, c_{2}=-2, \text { on comparing with } 3 x-6 y-2 z=7 \\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{(2)^{2}+(1)^{2}+(-2)^{2}} \cdot \sqrt{(3)^{2}+(-6)^{2}+(-2)^{2}}} \\ &=\frac{4}{3 \times 7}=\frac{4}{21} \\ &\theta=\cos ^{-1}\left(\frac{4}{21}\right) \end{aligned}$

The Plane exercise 28.6 question 3 (i)

Answer:
We need to show that the following planes are at right angle
Hint:
$\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=0$
Given:
$\overrightarrow{\boldsymbol{r}} \cdot(2 \hat{\boldsymbol{\imath}}-\hat{\boldsymbol{j}}+\widehat{\boldsymbol{k}})=\mathbf{5} \text { And } \vec{r} \cdot(-\hat{\imath}-\hat{\jmath}+\hat{k})=3$
Solution:
Here
$\begin{aligned} &\vec{n}_{1}=2 \hat{\imath}-\hat{\jmath}+\hat{k}, \text { on comparing with } \overrightarrow{\boldsymbol{r}} \cdot(\mathbf{\imath} \hat{\boldsymbol{\imath}}-\hat{\boldsymbol{j}}+\widehat{\boldsymbol{k}})=\mathbf{5}\\ &\vec{n}_{2}=-\hat{\imath}-\hat{\jmath}+\widehat{k} \text { on comparing wit } \vec{r} \vec{r} \cdot(-\hat{\imath}-\hat{\jmath}+\hat{k})=3\\ &\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}\\ &=\frac{(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(-\hat{\imath}-\hat{\jmath}+\hat{k})}{|2 \hat{\imath}-\hat{\jmath}+\hat{k}| \cdot|-\hat{\imath}-\hat{\jmath}+\hat{k}|}\\ &=\frac{-2+1+1}{\sqrt{(2)^{2}+(-1)^{2}+(1)^{2}} \cdot \sqrt{(-1)^{2}+(-1)^{2}+(1)^{2}}}\\ &=0 \end{aligned}$
$\theta =\frac{\pi }{2}$ .......hence proved

The Plane exercise 28.6 question 3 (ii)

Answer:
We need to show that the following planes are at right angle
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:
$x=2y+4z=10\hspace{0.2cm} And \hspace{0.2cm}18x+17y+4z=49$
Solution:
Here,
$\begin{aligned} &a_{1}=1, b_{1}=-2, c_{1}=4, \text { on comparing with } x-2 y+4 z=10\\ &a_{2}=18, b_{2}=17, c_{2}=4, \text { on comparing with } 18 x+17 y+4 z=49\\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\\ &=\frac{1(18)-2(17)+(4)(4)}{\sqrt{(1)^{2}+(-2)^{2}+(4)^{2}} \cdot \sqrt{(18)^{2}+(17)^{2}+(4)^{2}}}\\ &=0\\ &\theta=\frac{\pi}{2} \text {....hence proved } \end{aligned}$

The Plane exercise 28.6 question 4 (i)

Answer:
$\lambda =17$
Hint:
$\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot \vec{n}_{2} \mid}=0$
Given:
$\overrightarrow{\boldsymbol{r}} \cdot(\hat{\boldsymbol{\imath}}+2 \hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}})=7 \text { And } \vec{r} \cdot(\lambda \hat{\imath}+2 \hat{\jmath}-7 \hat{k})=26$
Solution:
Here
$\begin{aligned} &\vec{n}_{1}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \text { on comparing wit } \overrightarrow{\boldsymbol{r}} \cdot(\hat{\boldsymbol{\imath}}+2 \hat{\boldsymbol{j}}+3 \widehat{\boldsymbol{k}})=\mathbf{7}\\ &\vec{n}_{2}=\lambda \hat{\imath}+2 \hat{\jmath}-7 \hat{k}, \text { on comparing with } \vec{r} \cdot(\lambda \hat{\imath}+2 \hat{\jmath}-7 \hat{k})=26\\ &\vec{n}_{1} \cdot \vec{n}_{2}=0 \quad\left[\because\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right| \neq 0\right]\\ &(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \cdot(\lambda \hat{\imath}+2 \hat{\jmath}-7 \hat{k})=0\\ &\lambda+4-21=0\\ &\lambda=17 \end{aligned}$

The Plane exercise 28.6 question 4 (ii)

Answer:
2
Hint:
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=0$
Given:
$2x-4y+3z=5\hspace{0.2cm} And \hspace{0.2cm}x+2y+$ $\lambda z=5\\$
Solution:
Here
$\begin{aligned} &a_{1}=2, b_{1}=-4, c_{1}=3, \text { on comparing with } 2 x-4 y+3 z=5\\ &a_{2}=1, b_{2}=2, c_{2}=\lambda, \text { on comparing with } x+2 y+\lambda z=5\\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=0, \text { as t } \text { e planes are penpendicular } \cos \left(\frac{\pi}{2}\right)\\ &=0\\ &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\\ &2-8+3 \lambda=0\\ &\lambda=2 \end{aligned}$

The Plane exercise 28.6 question 4 (iii)

Answer:
$\lambda =0$
Hint:
$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$
Given:
$3x-6y-2z=7 And 2x+y-\lambda=5$
Solution:
Here
$\begin{aligned} &a_{1}=3, b_{1}=-6, c_{1}=-2, \text { on comparing with } 3 x-6 y-2 z=7\\ &a_{2}=2, b_{1}=1, c_{2}=-\lambda, \text { on comparing with } 2 x+y-\lambda z=5\\ &\text { Since the planes are perpendicular }\\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=0\\ &\therefore a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\\ &6-6+2 \lambda=0\\ &\lambda=0 \end{aligned}$

The Plane exercise 28.6 question 5

Answer:
Therefore, required equation of the plane is $5x+9y+11z-8=0$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
$\text { Point }(-1,-1,2) \text { and planes } 3 x+2 y-3 z=1 \text { and } 5 x-4 y+z=5$
Solution:
We know that equation of the plane passing through $\left ( x_{1},y_{1},z_{1} \right )$ is given as
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane passes through $\left ( -1,- 1,2\right )$ so the equation of the plane is
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
$ax+by+cz=2c-a-b$ (1)
Now the required plane is also perpendicular to the planes. $3x+2y-3z=1$ And
$5x-4y+z=5$
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1} =0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2} =0$ are at right angles
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have,
$3a+2b-3c=0$ (b)
$5a-4b+c=0$ (c)
Solving (b) and (c) we get
$\begin{aligned} &\frac{a}{2 \times 1-(-3) \times(-4)}=\frac{b}{5 \times(-3)-3 \times(1)}=\frac{c}{3 \times(-4)-2 \times(5)} \\ &\frac{a}{2-12}=\frac{b}{-15-3}=\frac{c}{-12-10} \\ &\frac{a}{-10}=\frac{b}{-18}=\frac{c}{-22}=\lambda \\ &a=-10 \lambda, b=-18 \lambda, c=-22 \lambda \end{aligned}$
Putting the values of a, b and c in equation (1) we have
$\begin{aligned} (-10 \lambda) x+(-18 \lambda) y+(-22 \lambda) z &=2(-22 \lambda)-(-10 \lambda)-(-18 \lambda) \\ -10 \lambda x-18 \lambda y-22 \lambda z &=-44 \lambda+10 \lambda+18 \lambda \\ -10 \lambda x-18 \lambda y-22 \lambda z &=-16 \lambda \end{aligned}$
Divide both sides by $\left ( -2\lambda \right )$ , we get
$5x+9y+11z=8$
So, the required of the plane equation is $5x+9y+11z-8=0$

The Plane exercise 28.6 question 6

Answer:
Therefore, required equation of the plane is $2x-4y+3z=8$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Point $\left ( 1,-3,-2 \right )$ and planes $x+2y+2z=5$ and $3x+3y+2z=8$
Solution:
We know that equation of the plane passing through $\left ( x_{1},y_{1},z_{1} \right )$ is given as,
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane passes through $\left ( 1,-3,-2 \right )$ so the equation of the plane is
$a\left ( x-1\right )+b\left ( y+3 \right )+c\left ( z+2 \right )=0$
$ax+by+cz=a-3b-2c$ (1)
Now the required plane is also perpendicular to the planes
$x+2y+2z=5$ And $3x+3y+2z=8$
We know that the plane $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are at right angles
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have,
$a+2b+2c=0$ (b)
$3a+3b+2c=0$ (c)
Solving (b) and (c) we get
$\begin{aligned} &\frac{a}{2 \times 2-3 \times 2}=\frac{b}{3 \times 2-1 \times 2}=\frac{c}{1 \times 3-2 \times 3} \\ &\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6} \\ &\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}=\lambda \\ &a=-2 \lambda, b=4 \lambda, c=-3 \lambda \end{aligned}$
Putting values of a, b and c in equation (1) we get
$\begin{aligned} &(-2 \lambda) x+(4 \lambda) y+(-3 \lambda) z=(-2 \lambda)-3(4 \lambda)-2(-3 \lambda) \\ &-2 \lambda x+4 \lambda y-3 \lambda z=-2 \lambda-12 \lambda+6 \lambda \\ &-2 \lambda x+4 \lambda y-3 \lambda z=-8 \lambda \end{aligned}$
Divide both sides by $\left ( -\lambda \right )$ we get
$2x-4y+3z=8$
So, the equation of the required planes is $2x-4y+3z=8$

The Plane exercise 28.6 question 7

Answer:
Therefore, required equation of the plane is $x+2y+5z=0$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Passing through the origin is $\left ( 0,0,0 \right )$ and planes $x+2y-z=1$ and $3x-4y+z=5$
Solution:
We know that solution of the plane passing through $\left ( x_{1} ,y_{1},z_{1}\right )$ is given as $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane passes through $\left ( 0,0,0 \right )$
So the equation of the plane is
$a\left ( x-0 \right )+b\left ( y-0 \right )+c\left ( z-0 \right )=0$
ax+by+cz=0 (1)
Now, the required plane is also perpendicular to the planes
$x+2y-z=1$ And $3x-4y+z=5$
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are at right angles
If $a_{1}a2+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we get
$a+2b-c=0$ (b)
$3a-4b+c=0$ (c)
Solving (b) and (c) we get
$\begin{aligned} &\frac{a}{2 \times 1-(-4)(-1)}=\frac{b}{3 \times(-1)-1 \times 1}=\frac{c}{1 \times(-4)-(2) \times 3} \\ &\frac{a}{2-4}=\frac{b}{-3-1}=\frac{c}{-4-6} \\ &\frac{a}{-2}=\frac{b}{-4}=\frac{c}{-10}=\lambda \\ &a=-2 \lambda, b=-4 \lambda, c=-10 \lambda \end{aligned}$

Putting values of a,b,c in equation (1) we get
$\begin{aligned} &(-2 \lambda) x-4 \lambda y-10 \lambda z=0 \\ &-2 \lambda x-4 \lambda y-10 \lambda z=0 \end{aligned}$
Divide both sides by $(-2 \lambda)$ we get
$x+2y+5z=0$
So the required equation of plane is
$x+2y+5z=0$

Plane exercise 28.6 question 8

Answer:
Therefore, required equation of the plane is $x+y-2z+4=0$
Hint:
Using the formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given:
Points $\left ( 1,-1,2 \right )$ and $\left ( 2,-2,2 \right )$ plane $6x-2y+2z=9$
Solution:
We know that solution of plane passing through $\left ( x,y,z \right )$ is given as,
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
The require plane passes through $\left ( 1,-1,2 \right )$
So, the equation of plane is
$a\left(x-1\right)+b\left(y+1\right)+c\left(z-2\right)=0$ (1)
Plane (1) is also passing through $\left ( 2,-2,2 \right )$ . So, $\left ( 2,-2,2 \right )$ must satisfy the equation of plane
So, we know
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
$a-b=0$ (2)
Plane $6x-2y+2z=9$ is perpendicular to the required plane
We know that plane $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are at right angle
If $a_{1}a2+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have
$a\left ( 6 \right )+b\left ( -2 \right )+c\left ( 2 \right )=0$
$a\left ( 6 \right )+b\left ( -2 \right )+c\left ( 2 \right )=0$ (3)
Solving (2) and (3) we get
$\begin{aligned} &\frac{a}{(-1) \times 2-(-2) \times 0}=\frac{b}{6 \times 0-1 \times 2}=\frac{c}{1 \times(-2)-6 \times(-1)} \\ &\frac{a}{-2-0}=\frac{b}{0-2}=\frac{c}{-2+6} \end{aligned}$
$\begin{aligned} &\frac{a}{-2}=\frac{b}{-2}=\frac{c}{4}=\lambda \\ &a=-2 \lambda, b=-2 \lambda, c=4 \lambda \end{aligned}$
Putting values of a, b and c in equation (1)
We get
$\begin{aligned} &(-2 \lambda)(x-1)+(-2 \lambda)(y+1)+(4 \lambda)(z-2)=0 \\ &-2 \lambda x+2 \lambda-2 \lambda y-2 \lambda+4 \lambda z-8 \lambda=0 \\ &-2 \lambda x-2 \lambda y+4 \lambda z-8 \lambda=0 \end{aligned}$
Divide by $\left ( -2\lambda \right )$ we get
$x+y-2z+4=0$
So, the required plane is $x+y-2z+4=0$

The Plane exercise 28.6 question 9

Answer:
Therefore, required equation of the plane is $3x+4y-5z=9$
Hint:
Using the formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given:
Points $\left ( 2,2,1\right )$ and $\left ( 9,3,6\right )$ plane $3x+4y-5z=9$
Solution:
We know that solution of a plane passing through $\left ( x_{1} ,y_{1},z_{1}\right )$ is given as,
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
The required plane passing through $\left ( 2,2,1 \right )$
So the equation of the plane is
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ (1)
Plane (1) is also passing through $\left ( 9,3,6 \right )$
So $\left ( 9,3,6 \right )$ must satisfy the equation of plane
$a\left ( 9-2 \right )+b\left ( 3-2 \right )+c\left ( 6-1 \right )=0$
$7a+b+5c=0$ (2)
Plane $2x+6y+6z=1$ is perpendicular to required plane
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are the right angle if
$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have
$2a+6b+6c=0$ (3)
Solving (2) and (3) we get
$\begin{aligned} &\frac{a}{1 \times 6-5 \times 6}=\frac{b}{2 \times 5-7 \times 6}=\frac{c}{7 \times 6-2 \times 1} \\ &\frac{a}{6-30}=\frac{b}{10-42}=\frac{c}{42-2} \\ &\frac{a}{-24}=\frac{b}{-32}=\frac{c}{40}=\lambda \\ &a=-24 \lambda, b=-32 \lambda, c=40 \lambda \end{aligned}$

Putting the values of a, b and c in equation (1)
$\begin{aligned} &(-24 \lambda)(x-2)+(-32 \lambda)(y-2)+(40 \lambda)(z-1)=0 \\ &-24 \lambda x+48 \lambda-32 \lambda y+64 \lambda+40 \lambda z-40 \lambda=0 \\ &-24 \lambda x-32 \lambda y+40 \lambda z+72 \lambda=0 \end{aligned}$
Divide by $\left ( -8\lambda \right )$ we get
$3x+4y-5z-9=0$
So, required plane is $3x+4y-5z-9=0$

The Plane exercise 28.6 question 10

Answer:
Therefore, required equation of the plane is $2x+2y+-3x+3=0$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Points $\left ( -1,1,1 \right )$ and $\left ( 1,-1,1 \right )$ plane $x+2y+2z=5$
Solution:
We know that solution of a plane passing through $\left ( x_{1} ,y_{1},z_{1}\right )$ is given as
$a\left ( x+x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane is passing through $\left ( -1,1,1 \right )$
So, the equation of the plane is,
$a\left ( x+1 \right )+b\left ( y-1 \right )+c\left ( z-1 \right )=0$ (1)
Plane (1) also passing through $\left ( 1,-1,1 \right )$
So $\left ( 1,-1,1 \right )$ must satisfy the equation of plane
$a\left ( 1+1 \right )+b\left ( -1-1 \right )+c\left ( 1-1 \right )=0$
$2a-2b=0$ (2)
Plane $x+2y+2z=5$ is perpendicular to the required plane
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are right angle
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have
$a\left ( 1 \right )+b\left ( 2 \right )+c\left ( 2 \right )=0$
$a+2b+2c=0$ (3)
Solving (2) and (3) we get
$\begin{gathered} \frac{a}{(-2) \times 2-2 \times 0}=\frac{b}{1 \times 0-2 \times 2}=\frac{c}{2 \times 2-1 \times(-2)} \\ \frac{a}{-4-0}=\frac{b}{0-4}=\frac{c}{4+2} \\ \frac{a}{-4}=\frac{b}{-4}=\frac{c}{6}=\lambda \\ a=-4 \lambda, b=-4 \lambda, c=6 \lambda \end{gathered}$
Putting the value of a, b and c in equation (1) we get
$\begin{aligned} &(-4 \lambda)(x+1)+(-4 \lambda)(y-1)+6 \lambda(z-1)=0 \\ &-4 \lambda x-4 \lambda-4 \lambda y+4 \lambda+6 \lambda z-6 \lambda=0 \\ &-4 \lambda x-4 \lambda y+6 \lambda z-6 \lambda=0 \end{aligned}$
Dividing by $\left ( -2\lambda \right )$ we get
$2x+2y-3z+3=0$
So the required plane is $2x+2y-3z+3=0$

The Plane exercise 28.6 question 11

Answer:
Therefore, required equation of the plane is $y=3$
Hint:
Using the properties of plane
Given:
Plane with intercept 3 on the y-axis and parallel to ZOX plane
Solution:
We know that the equation of ZOX plane will is y=0. So, a plane parallel to plane ZOX will have the equation
y=constant
Now, it is given that the plane makes an intercept of 3 on y-axis. So, the value of constant is equal to 3
Therefore, the required equation of plane is y=3.

The Plane exercise 28.6 question 12

Answer:
Therefore, required equation of the plane is $5x-4y-z=7$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Points $\left ( 1,-1,2 \right )$
Plane $2x+3y-2z=5$ and $x+2y-3z=8$
Solution:
We know that solution of a plane passing through $\left ( x_{1},y_{1},z_{1}\right )$ is given as
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane passes through $\left ( 1,-1,2 \right )$ .
So the equation plane is $a\left ( x-1\right )+b\left ( y+1 \right )+c\left ( z-2 \right )=0$
$ax+by+cz=a-b+2c$ (1)
Now the required plane is also perpendicular to the planes,
$2x+3y-2z=5$ & $x+2y-3z=8$
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ & $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are the right angle
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we get
$2a+3b-2c=0$ (b)
$a+2b-3c=0$ (c)
Solving (b) and (c) we get
$\begin{aligned} &\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ &\frac{a}{-9+4}=\frac{b}{-2+6}=\frac{c}{4-3} \\ &\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}=\lambda \\ &a=-5 \lambda, b=4 \lambda, c=\lambda \end{aligned}$
Putting values of a, b and c in equation (1) we get
$\begin{aligned} &\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ &\frac{a}{-9+4}=\frac{b}{-2+6}=\frac{c}{4-3} \\ &\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}=\lambda \\ &a=-5 \lambda, b=4 \lambda, c=\lambda \end{aligned}$
Dividing both the sides by $\left ( -\lambda \right )$ , we get
$5x-4y-z=7$
So, the required plane is $5x-4y-z=7$

The Plane exercise 28.6 question 13

Answer:
Therefore, required equation of the plane is $x+y+z=a+b+c$
Hint:
Using properties of plane
Given:
Points $\left ( a,b,c \right )$ and plane $\overrightarrow{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )=2$
Solution:
The required plane is parallel to the plane $\overrightarrow{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )=2$
Any plane parallel to $\overrightarrow{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )=2$ is given as $\overrightarrow{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )=k$
Further, it is given that the plane is passing through $\left ( a,b,c \right )$ . So point $\left ( a,b,c \right )$ should satisfy the equation of the plane
We have $\left ( a\hat{i}+b\hat{j}+c\hat{k} \right )\cdot \left (\hat{i}+\hat{j}+\hat{k} \right )=k$
$a+b+c=k$
Hence, the equation of the required plane is,
$\overrightarrow{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )= a+b+c$
Or, $x+y+z= a+b+c$

The Plane exercise 28.6 question 14

Answer:
Therefore, required equation of the plane is $7x-8y+3z+25=0$
Hint:
Using the formula $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Points $\left ( -1,3,2 \right )$
Plane $x+2y+3z=5$ and $3x+3y+z=0$
Solution:
We know that solution of a plane passing through $\left ( x_{1} ,y_{1},z_{1}\right )$ are given as
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The required plane passes through
So, the equation of plane is
$a\left ( x+1 \right )+b\left ( y-3 \right )+c\left ( z-2 \right )=0$
$ax+by+cz=3b+2c-a$ (1)
Now, the required plane is also perpendicular to the planes,
$x+2y+3z=5$ & $3x+3y+z=0$
We know that planes $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ & $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are right angle
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we get
$a+2b+3c=0$ (b)
$3a+3b+c=0$ (c)
Solving (b) and (c) we get
$\begin{aligned} &\frac{a}{2 \times 1-3 \times 3}=\frac{b}{3 \times 3-1 \times 1}=\frac{c}{1 \times 3-3 \times 2} \\ &\frac{a}{2-9}=\frac{b}{9-1}=\frac{c}{3-6} \\ &\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=\lambda \\ &a=-7 \lambda, b=8 \lambda, c=-3 \lambda \end{aligned}$
Putting values of a, b and c in equation (1) we get
$\begin{aligned} &(-7 \lambda) x+(8 \lambda) y+(-3 \lambda) z=3(8 \lambda)+2(-3 \lambda)+7 \lambda \\ &-7 \lambda x+8 \lambda y-3 \lambda z=24 \lambda-6 \lambda+7 \lambda \\ &-7 \lambda x+8 \lambda y-3 \lambda z=25 \lambda \end{aligned}$
Dividing both $\left ( -\lambda \right )$ we get
$7x-8y+3z+25=0$

The Plane exercise 28.6 question 15

Answer:
Therefore, required equation of the plane is $18x+17y+4z=49$ $a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Hint:
Using the formula
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
Given:
Points $\left ( 2,1,-1 \right )$ and $\left ( -1,3,4 \right )$
Plane $x-2y+4z=10$
Solution:
We know that solution of plane passing through $\left ( x_{1},y_{1} ,z_{1}\right )$ is given as,
$a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=0$
The require plane passes through $\left ( 2,1,-1 \right )$
So, the equation of plane is
$a\left ( x-2 \right )+b\left ( y-1 \right )+c\left ( z+1 \right )=0$ (1)
Plane (1) is also passing through $\left ( -1,3,4 \right )$ . So, $\left ( -1,3,4 \right )$ must satisfy the equation of plane
So, we know
$a\left ( -1-2 \right )+b\left ( 3-1 \right )+c\left ( 4+1 \right )=0$
$-3a+2b+5c=0$ (2)
Plane $x-2y+4z=10$ is perpendicular to the required plane
We know that plane $a_{1}x+b_{1}y+c_{1}z+d_{1}=0$ and $a_{2}x+b_{2}y+c_{2}z+d_{2}=0$ are at right angle
If $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$ (a)
Using (a) we have
$a-2b+4c=0$ (3)
Solving (2) and (3) we get
$\begin{aligned} &\frac{a}{4 \times 2+5 \times 2}=\frac{b}{5 \times 1+3 \times 4}=\frac{c}{3 \times 2-2 \times 1} \\ &\frac{a}{18}=\frac{b}{17}=\frac{c}{4} \\ &\frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda \\ &a=18 \lambda, b=17 \lambda, c=4 \lambda \end{aligned}$
Putting values of a, b and c in equation (1)
We get
$\begin{aligned} &18 \lambda(x-2)+17 \lambda(y-1)+4 \lambda(z+1)=0 \\ &18 \lambda x-36 \lambda+17 \lambda y-17 \lambda+4 \lambda z+4 \lambda=0 \\ &18 \lambda x+17 \lambda y+4 \lambda z-49 \lambda=0 \end{aligned}$
Divide by $\left ( \lambda \right )$ we get
$18x+17y+4z-49=0$
So, the required plane is
$18x+17y+4z-49=0$

The Class 12 RD Sharma chapter 28 exercise 28.6 solution covers the chapter 'The Plane.' There are about 24 questions in this exercise that are extremely basic and simple. The RD Sharma class 12th exercise 28.6 covers all the essential concepts of this chapter that are mentioned below,

Equation of plane
Vector equation of plane
Equation of plane passing through points
Equation of line under planes condition
Angles between the two planes
Equation of plane passing through the point and perpendicular to the plane.

Mentioned below are few benefits of the RD Sharma class 12th exercise 28.6 :-

The RD Sharma class 12 chapter 28 exercise 28.6 is used by teachers to assign homework therefore practice from this solution can help you solve homework easily.
The questions given in the RD Sharma class 12th exercise 28.6 are often similar to the ones that are asked in the board exams so, if you keep a thorough practice of the book then it's an easy task for you to score high in exams.
The RD Sharma class 12 solution chapter 28 exercise 28.6 is updated every year to the latest version so that no concept is missed out for students to practice.
The RD Sharma solutions are of the same syllabus that is of the NCERT and therefore it is useful for considering it for the practice for public exams as well.
Many solved questions and useful tips are provided in the solution to take in reference when having a problem in solving any question.
The online study material and PDFs of the RD Sharma class 12th exercise 28.6 is available on the Career360 website and you don't have to pay any amount for downloading these solutions.

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Frequently Asked Questions (FAQs)

1. What is a plane?

A plane is a flat, 2-dimensional surface having infinite dimensions but for having zero thickness. Such that if any two points are taken on it, the line segment joining them lies completely on the surface

2. What is the equation of a 3-dimensional plane?

A plane in 3-dimensional space has the following equation

ax + by + cz + d = 0, where at least one of the values of a, b, c, d, must be a non-zero.

3. What are the three possible planes in the 3-D coordinate system?

They are,

XY plane where the value of z coordinates is zero

YZ plane, where the value of x coordinates is zero

ZX plane, where the value of y coordinates is zero

4. Is the RD Sharma class 12 chapter 28 of the latest version?

Yes, these solutions are regularly updated to correspond with the syllabus of NCERT textbooks, which helps crack public examinations.

5. From where can I download the RD Sharma class 12 chapter 28 solution?

You can download the E-book and PDF from the Career360 website that also free of cost.