RD Sharma Class 12 Exercise 28.8 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.8 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:55 AM IST

RD Sharma is a well-known name for many students in CBSE schools. Many schools widely use the RD Sharma class 12 solution of The plane exercise 28.8 for maths. Not only students, the RD Sharma class 12th exercise 28.8 is used by many teachers and experts to take in reference while giving lectures or preparing question papers. Class 12 RD Sharma chapter 28 exercise 28.7 solution is well known for its basic understanding and simple concepts which are very easy for beginners. RD Sharma solutions Their main motive is to encourage students to learn maths in the simplest possible way.

## The Plane Excercise: 28.8

The Plane exercise 28.8 question 1

Answer:- The answer of the given question is $2x-3y+z=7$.
Hint:- By find and putting the value of $k$
Given:-$2x-3y+z=0$ and point $(1,-1,2)$
Solution:- Given equation of plane is $2x-3y+z=0$ … (i)
We know that equation of a plane parallel to given plane (i) is
$2x-3y+z+k=0$ …(ii)
As given that, plane (ii) is passing through the point $(1,-1,2)$ so it satisfy the plane (ii)
$\begin{gathered} 2(1)-3(-1)+2+k=0 \\\\ k=-7 \end{gathered}$
Put the value of k in equation (ii),
$2x-3y+z-7=0$
So, equation of the required plane is, $2x-3y+z=7$

The Plane exercise 28.8 question 2

Answer:- The answer of the given question is $\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$
Hint:- By putting the value of k in equation (ii)
Given:- $\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+2=0$
Solution:- Given equation of a plane is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+2=0$ …(i)
We know that the equation of a plane parallel to given plane (i) is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+k=0$ … (ii)
As given that plane (ii) is passing through the point $3 \hat{\imath}+4 \hat{\jmath}-\hat{k}$ so it satisfy the equation (ii)
\begin{aligned} &(3 \hat{\imath}+4 \hat{\jmath}-\hat{k}) \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+k=0 \\\\ &(3)(2)+(4)(-3)+(-1)(5)+k=0 \end{aligned}
$k=11$
Put the value of k in equation (ii)
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$
So, the equation of the required plane is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$

The Plane exercise 28.8 question 3

Answer:- The answer of the given question is $15x-47y+28z-7=0.$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x-7y+4z-3=0$ and $3x-5y+4z+11=0$
Solution:- So equation of plane passing through the line of intersection of given two planes is
\begin{aligned} &(2 x-7 y+4 z-3)+k(3 x-5 y+4 z+11)=0 \\\\ &2 x-7 y+4 z-3+3 k x-5 k y+4 k z+11 k=0 \\\\ &x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \end{aligned} … (i)
As given that, plane (i) is passing through the point $(-2,1,3)$ so it satisfy the equation (i)
$(-2)(2+3 k)+(1)(-7-5 k)+(3)(4+4 k)-3+11 k=0$
$k=\frac{1}{6}$
Put the value of k in equation (i)
$\begin{gathered} x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \\\\ x\left(2+\frac{3}{6}\right)+y\left(-7-\frac{5}{6}\right)+z\left(4+\frac{4}{6}\right)-3+\frac{11}{6}=0 \end{gathered}$
$\begin{gathered} x\left(\frac{12+3}{6}\right)+y\left(\frac{-42-5}{6}\right)+z\left(\frac{24+4}{6}\right)-\frac{18+11}{6}=0 \\\\ x\left(\frac{15}{6}\right)+y\left(-\frac{47}{6}\right)+z\left(\frac{28}{6}\right)-\frac{7}{6}=0 \end{gathered}$
Multiplying by 6 we get
$15 x-47 y+28 z-7=0$

The Plane exercise 28.8 question 4

Answer:- The answer of the given question is $\vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0$
Hint:-$\vec{r} \cdot\left(\overrightarrow{n_{1}}+k \overrightarrow{n_{2}}\right)=d_{1}+k d_{2}$
Given:-$\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath}-\hat{k})=0$ and $\vec{r} \cdot(\hat{\jmath}+2 \hat{k})=0$ , Point $(2 \hat{\imath}+\hat{j}-\hat{k})$
Solution:- $\vec{r} \cdot \overrightarrow{n_{1}}=d_{1}$
$\vec{r} \cdot \overrightarrow{n_{2}}=d_{2}$ is given by
$\vec{r}\left(\overrightarrow{n_{1}}+k \overrightarrow{n_{2}}\right)=d_{1}+k d_{2}$
∴ The equation of the plane passing through the line of intersection of given two planes
$\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath}-\hat{k})=0$ and $\vec{r} \cdot(\hat{j}+2 \hat{k})=0$ is given by
$\vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(\hat{j}+2 \hat{k})\}=0$ … (i)
As given that, plane (i) is passing through the point $2 \hat{\imath}+\hat{\jmath}-\hat{k}$
$\begin{gathered} (2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\jmath}+2 \hat{k})=0 \\\\ (2)(1)+(1)(3)+(-1)(-1)+k[(2)(0)+(1)(1)+(-1)(2)]=0 \end{gathered}$
$\begin{gathered} (2+3+1)+k(1-2)=0 \\\\ 6-k=0 \\\\ k=6 \end{gathered}$
Putting the value of k in equation (i)
$\begin{gathered} \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(\hat{\jmath}+2 \hat{k})\}=0 \\\\ \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+6(\hat{\jmath}+2 \hat{k})\}=0 \\\\ \vec{r} \cdot(\hat{\imath}+9 \hat{j}+11 \hat{k})=0 \end{gathered}$
So, the equation of required plane is
$\vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0$

The Plane exercise 28.8 question 5

Answer:- The answer of the given question is $28x-17y+9z=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x-y=0$ and $3z-y=0$ and plane $4x+5y-3z=8$
Solution:- So equation of plane passing through the line of intersection of given two planes is
$2x-y=0$ and $3z-y=0$ is
$\begin{gathered} (2 x-y)+k(3 z-y)=0 \\\\ 2 x-y+3 k z-k y=0 \end{gathered}$
$x(2)+y(-1-k)+z(3 k)=0$ …(i)
We know that, two planes are perpendicular if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …(ii)
Given, plane (i) is perpendicular to plane
$4x+5y-3z=8$ …(iii)
Using (i) and (iii) in equation (ii)
$\begin{gathered} (2)(4)+(-1-k)(5)+(3 k)(-3)=0 \\\\ 3-14 k=0 \end{gathered}$
$k=\frac{3}{14}$
Putting value of k in equation (i)

$\begin{gathered} x(2)+y(-1-k)+z(3 k)=0 \\\\ x(2)+y\left(-1-\frac{3}{14}\right)+z\left(3\left(\frac{3}{14}\right)\right)=0 \\\\ x(2)+y\left(-\frac{17}{14}\right)+z\left(\frac{9}{14}\right)=0 \end{gathered}$
Multiplying with 14 we get
$28x-17y+9z=0$

The Plane exercise 28.8 question 6

Answer:- The answer of the given question is $33x+45y+50z-41=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+2y+3z-4=0$ and
$2x+y-z+5=0$
Solution:- So equation of plane passing through the line of intersection of given two planes
$x+2y+3z-4=0$ and$2x+y-z+5=0$ is given by
\begin{aligned} &(x+2 y+3 z-4)+k(2 x+y-z+5)=0 \\\\ &x+2 y+3 z-4+2 k x+k y-k z+5 k=0 \end{aligned}
$x(1+2 k)+y(2+k)+z(3-k)-4+5 k=0$ … (i)
We know that, two planes are perpendicular if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ … (ii)
Given, plane (i) is perpendicular to plane,
$5x+3y-6z+8=0$ … (iii)
Using (i) and (iii) in equation (ii)
$\begin{gathered} 5(1+2 k)+3(2+k)+(-6)(3-k)=0 \\\\ 5+10 k+6+3 k-18+6 k=0 \\\\ k=\frac{7}{19} \end{gathered}$
Putting the value of k in equation (iii)
$\begin{gathered} x\left(1+\frac{14}{19}\right)+y\left(2+\frac{7}{19}\right)+z\left(3-\frac{7}{19}\right)-4+\frac{35}{19}=0 \\\\ x \frac{(19+14)}{19}+y\left(\frac{38+7}{19}\right)+z\left(\frac{57-7}{19}\right)+\frac{-76+35}{19}=0 \end{gathered}$
$x\left(\frac{33}{19}\right)+y\left(\frac{45}{19}\right)+z\left(\frac{50}{19}\right)-\frac{41}{19}=0$
Multiplying with 19 we get
$33x+45y+50z-41=0$

The Plane exercise 28.8 question 7

Answer:- The answer of the given question is $x-10y-5z=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+2y+3z+4=0$ and $x-y+z+3=0$
Solution:- So equation of plane passing through the line of intersection of given two planes
$x+2y+3z+4=0$ and $x-y+z+3=0$
$\begin{array}{r} (x+2 y+3 z+4)+k(x-y+z+3)=0 \\\\ x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \end{array} .$ … (i)
Equation (i) is passing through origin, so
$\begin{gathered} (1+k)+(0)(2-k)+(0)(3+k)+4+3 k=0 \\\\ 0+0+0+4+3 k=0 \\\\ k=-\frac{4}{3} \end{gathered}$

Put the value of k in equation (i)

$\begin{gathered} x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(2+\frac{4}{3}\right)+z\left(3-\frac{4}{3}\right)+4-\frac{12}{3}=0 \end{gathered}$
$\begin{gathered} x\left(\frac{3-4}{3}\right)+y\left(\frac{6+4}{3}\right)+z\left(\frac{9-4}{3}\right)+4-4=0 \\\\ -\frac{x}{3}+\frac{10 y}{3}+\frac{5 z}{3}=0 \end{gathered}$
Multiplying by 3, we get
$-x+10y+5z=0$
$x-10y-5z=0$

The Plane exercise 28.8 question 8

Answer:- The answer of the given question is $\vec{r}(\hat{\imath}+7 \hat{\jmath})+13=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x-3y+2z-5=0$ and $2x-y+3z-1=0$
Solution:- So equation of plane passing through the line of intersection of planes
$x-3y+2z-5=0$ and $2x-y+3z-1=0$ is given by
\begin{aligned} (x-3 y+2 z-5) &+k(2 x-y+3 z-1) &=0 & \\\\ x(1-2 k)+y(-3-k)+z(2+3 k)-5-k &=0 & \end{aligned} . … (i)
Plane (i) is passing the through the point$(1,-2,3),$ so

$\begin{gathered} 1(1+2 k)+(-2)(-3-k)+(3)(2+3 k)-5-k=0 \\\\ 1+2 k+6+2 k+6+9 k-5-k=0 \\\\ 8+12 k=0 \\\\ k=-\frac{2}{3} \end{gathered}$
Putting the value of k in equation (i)
$\begin{gathered} x(1+2 k)+y(-3-k)+z(2+3 k)-5-k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(-3+\frac{2}{3}\right)+z\left(2-\frac{6}{3}\right)-\left(\frac{15-2}{3}\right)=0 \\\\ -\frac{1}{3} x-\frac{7}{3} y-\frac{13}{3}=0 \end{gathered}$
Multiplying by -3
$x+7y+13=0$
$\begin{gathered} (x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}+7 \hat{\jmath})+13=0 \\\\ \vec{r} \cdot(\hat{\imath}+7 \hat{\jmath})+13=0 \end{gathered}$

The Plane exercise 28.8 question 9

Answer:- The answer of the given question is $51x+15y-50z+173=0.$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$5x+3y+6z+8=0$
$(x +2y+3z-4)=0$ and
$x +2y+3z-4=0$
Solution:- The equation of a plane through the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$
\begin{aligned} (x+2 y+3 z-4) &+\lambda(2 x+y-z+5)=0 \\\\ x(1+2 \lambda)+y(2+\lambda)+z(-\lambda+3)-4+5 \lambda &=0 \end{aligned} . … (i)

Also this is perpendicular to the plane $5x+3y+6z+8=0$ …(ii)
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

$\begin{gathered} 5(1+2 \lambda)+3(2+\lambda)+6(-\lambda+3)=0 \\\\ \therefore 5+10 \lambda+6+3 \lambda+18-6 \lambda=0 \\\\ \lambda=-\frac{29}{7} \end{gathered}$
∴ Putting this value of $\lambda$ in equation (i) we get equation of plane as
$51x+15y-50z+173=0$

The Plane exercise 28.8 question 10

Answer:- The answer of the given question is $\vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})+6=0$ or
$\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})+6=0$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$

Given:-\begin{aligned} &-\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6=0 \end{aligned}
$\\\\ \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})=0$
Solution:- The equation of a plane through the line of intersection of the planes$r \cdot(\hat{\imath}+3 \hat{j})+6=0$ and $\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 k)=0$
$\begin{gathered} \vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6+\lambda[\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})]=0 \\\\ \vec{r} \cdot[(\hat{\imath}+3 \hat{\jmath})]+6+3 \lambda \cdot \vec{r} \hat{\imath}-\vec{r} \lambda \hat{j}-4 \lambda \vec{r} \hat{k}=0 \end{gathered}$
$r(\hat{\imath}+3 \hat{\jmath}+3 \lambda \hat{\imath}-\lambda \hat{\jmath}-4 \lambda \hat{k})=-6$ ...........(i)
$\vec{r} \cdot \frac{[\hat{\imath}(1+3 \lambda)+\hat{j}(3-\lambda)+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}$
The perpendicular distance from the origin is unity
\begin{aligned} &\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1 \\\\ &(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36 \end{aligned}
$\begin{gathered} 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36 \\\\ \lambda^{2}=1 \\\\ \lambda=\pm 1 \end{gathered}$
Using equation (i) the required plane is
$\vec{r} \cdot(1 \pm 3) \hat{\imath}+(3 \mp 1) \hat{\jmath}+(\mp 4) \hat{k}=-6$
$\vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})=-6 \text { or }\vec{r} \cdot(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})=-6$
$\vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+6=0 \text { orr. }(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})+6=0$

The Plane exercise 28.8 question 11

Answer:- The answer of the given question is $7x+13y+4z=9$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x+3y-z+1=0$ and $x+y-2z+3=0$
Plane $:3x-y-2z-4=0$
Solution:- Cartesian form of equation of plane through the line of intersection of plane is
$A_{1} x+B_{1} y+C_{1} z+D_{1}+\lambda\left(A_{2} x+B_{2} y+C_{2} z+D_{2}\right)=0$ … (i)
Here the standard equation of plane is $A_{1} x+B_{1} y+C_{1} z+D_{1} \text { and } A_{2} x+B_{2} y+C_{2} z+D_{2}$
Substituting the values in equation (i) we get
$2 x+3 y-z+1+\lambda(x+y-2 z+3)=0$
$(2+\lambda) x+(3+\lambda) y+(-1-2 \lambda) z+1+3 \lambda=0$ ...........(ii)
It is given that the plane $3x-y-2z-4=0$ is perpendicular to the plane
We know that $\emptyset=90^{\circ}$ where $\cos 90^{\circ}=0$ so we get
$A_{1} A_{2}+B_{1} B_{2}+C_{1} C_{2}=0$ … (iii)
By comparing the standard equation in Cartesian form$\begin{array}{lll} A_{1}=2+\lambda, & B_{1}=3+\lambda, & C_{1}=-1-2 \lambda \\\\ A_{2}=3, & B_{2}=-1, & C_{2}=-2 \end{array}$
Substituting these values in equation (iii)
$(2+\lambda) \cdot 3+(3+\lambda)(-1)+(-1-2 \lambda)(-2)=0$
On further calculation
$\begin{gathered} 6+3 \lambda-3-\lambda+2+4 \lambda=0 \\ \lambda=-\frac{5}{6} \end{gathered}$
Substituting the value of λ in equation (ii)
$\left(2+\frac{-5}{6}\right) x+\left(3+\frac{-5}{6}\right) y+\left(-1-2 \cdot \frac{-5}{6}\right) z+1+3 \cdot \frac{-5}{6}=0$
By taking LCM
$\left(\frac{12-5}{6}\right) x+\left(\frac{18-5}{6}\right) y+\left(\frac{-6+10}{6}\right) z+\frac{6-15}{6}=0$
We get
$7x+13y+4z-9=0$
$7x+13y+4z=9$

The Plane exercise 28.8 question 12

Answer:- The answer of the given question is $33x+45y+50z-41=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:- $\vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4=0$ … (i)
$\vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5=0$ … (ii)
Solution:- The equation of the plane passing through the line of intersection of the plane given in equation (i) and equation (ii) is
$[\vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4]+\lambda[\vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5]=0$
$\vec{r}[(2 \lambda+1) \hat{\imath}+(\lambda+2) \hat{\jmath}+(3-\lambda) \hat{k}]+(5 \lambda-4)=0$ …(iii)
The plane in equation (iii) is perpendicular to the plane,
$\vec{r} \cdot(5 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})+8=0$
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

$\begin{gathered} \therefore 5(2 \lambda+1)+3(\lambda+2)-6(3-\lambda)=0 \\\\ 19 \lambda-7=0 \\\\ \lambda=\frac{7}{19} \end{gathered}$
Substituting $\lambda=\frac{7}{19}$ in equation (iii), we obtain
$\vec{r} \cdot\left[\frac{33}{19} \hat{\imath}+\frac{45}{19} \hat{\jmath}+\frac{50}{19} \hat{k}\right]-\frac{41}{19}=0$
$\vec{r} \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0$ … (iv)
This is vector equation of the required plane
Now $(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0$
$33x+45y+50z-41=0$

The Plane exercise 28.8 question 13

Answer:- The answer of the given question is $\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6$ and
$\vec{r} \cdot\left(2 \hat{\imath}+\widehat{3}_{j}+4 \hat{k}\right)=-5$ and Point $(1,1,1)$
Solution:- The Cartesian equation of the given planes are $x+y+z-6=0$ and
$2x+3y+4z+5=0$
The family of planes is$x+y+z-6+\lambda(2 x+3 y+4 z+5)=0$ …(i)
Since it pass through $(1,1,1)$
$\begin{gathered} 1+1+1-6+\lambda(2+3+4+5)=0 \\\\ -3+\lambda(14)=0 \\ \lambda=\frac{3}{14} \end{gathered}$
$\begin{gathered} x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0 \\\\ 14 x+14 y+14 z-84+6 x+9 y+12 z+15=0 \\\\ 20 x+23 y+26 z-69=0 \end{gathered}$
Vector equation is $\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69$

The Plane exercise 28.8 question 14

Answer:- The answer of the given question is $\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$
Hints:- We know the line of intersection of the plane $\vec{r} \cdot \vec{n}_{1}-\vec{d}_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-\vec{d}_{2}=0$ is
given by $\vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0$
Given :-The intersection of the plane vector
\begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})=7 \quad \text { and } \\\\ &\vec{r} \cdot(2 \hat{i}+3 \hat{j}+3 \hat{k})=9 \end{aligned}
Solution:-We know that, the equation of a plane through the line of intersection of the plane
$\vec{r} \cdot \vec{n}_{1}-d_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-d_{2}=0$
Is given by $\vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0$
So, equation of plane passing through the line of intersection of plane
\begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7=0 \text { and } \\ &\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0 \end{aligned}is given by
\begin{aligned} &{[\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7]+k[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]=0} \\\\ &\vec{r}[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \end{aligned}.............(1)
Given that plane (1) is passing through $2 \hat{i}+\hat{j}+3 \hat{k} \text { so }$
\begin{aligned} &(2 \hat{i}+\hat{j}+3 \hat{k})[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\\\ &(2)(2+2 k)+(1)(1+5 k)+(3)(3+3 k)-7-9 k=0 \\\\ &4+4 k+1+5 k+9+9 k-7-9 k=0 \end{aligned}
\begin{aligned} &9 k=-7 \\\\ &k=\frac{-7}{9} \end{aligned}
Put the value of k in equation (1)
\begin{aligned} &\vec{r}[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\\\ &\vec{r}\left[\left(2-\frac{14}{9}\right) \hat{i}+\left(1-\frac{35}{9}\right) \hat{j}+\left(3-\frac{21}{9}\right) \hat{k}\right]-7+\frac{63}{9}=0 \end{aligned}
\begin{aligned} &\vec{r}\left[\left(\frac{18-14}{9}\right) \hat{i}+\left(\frac{9-35}{9}\right) \hat{j}+\left(\frac{27-21}{9}\right) \hat{k}\right]-7+7=0 \\\\ &r\left[\left(\frac{4}{9}\right) \hat{i}-\left(\frac{26}{9}\right) \hat{j}+\left(\frac{6}{9}\right) \hat{k}\right]=0 \end{aligned}
Multiplying by $(\frac{9}{2})$,we get
$\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$
Equation of the required plane is
$\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$

The Plane exercise 28.8 question 15

Answer:- The answer of the given question is $7x-5y+4z-8=0.$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$3x-y+2z=4$ and $x+y+z=2$ , Point $(2,2,1)$
Solution:- The equation of any plane through the intersection of the planes is
$3x-y+2z-4=0$ and $x+y+z-2=0$ is
$(3 x-y+2 z-4)+\alpha(x+y+z-2)=0$, where $\alpha \in R$ …(i)
The plane passes through the point $(2, 2,1).$ Therefore, this point will satisfy equation (1)
$\begin{gathered} \therefore(3 \times 2-2+2 \times 1-4)+\alpha(2+2+1-2)=0 \\\\ 2+3 \alpha=0 \\ \alpha=-\frac{2}{3} \end{gathered}$
Substituting $\alpha=-\frac{2}{3}$ in equation (i), we obtain
$\begin{gathered} (3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0 \\\\ 3(3 x-y+2 z-4)-2(x+y+z-2)=0 \\\\ (9 x-3 y+6 z-12)-2(x+y+z-2)=0 \\\\ 7 x-5 y+4 z-8=0 \end{gathered}$

The Plane exercise 28.8 question 16

Answer:- The answer of the given question is$\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0 \text {. }$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:- $x+y+z=1$ ,
$2x+3y+4z=5$
$x-y+z=0$
Solution:- Equation of the plane through the intersection of plane is
\begin{aligned} &(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \text { or } \\\\ &(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-(1+5 \lambda)=0 \end{aligned} …(i)
This plane is perpendicular to $x-y+z=0$
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

$\begin{array}{r} 1(1+2 \lambda)-1(1+3 \lambda)+1(1+4 \lambda)=0 \text { or } \\\\ \qquad \lambda=-\frac{1}{3} \end{array}$
Equation of plane is
$\begin{gathered} (x+y+z-1)-\frac{1}{3}(2 x+3 y+4 z-5)=0 \\\\ \Rightarrow x-z+2=0 \end{gathered}$
Vector form of plane is $\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0$
Yes, line lies on the plane on $(-2, 3, 0)$ satisfies$\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0$

The Plane exercise 28.8 question 17

Answer:- The answer of the given question is $x+y+z=a+b+c$.
Hints:- By substituting $\vec{r}=x \hat{\imath}+y \hat{j}+z \hat{k}$ in equation (ii)
Given:-$(a, b, c)$ and parallel to plane $\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2$
Solution:- Any plane passes through the point $(a, b, c)$ and parallel to plane $\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2$ is given by
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=\lambda$…(i)
Here, the position vector $\vec{r}$ of this point is $\vec{r}=a \hat{\imath}+b \hat{\jmath}+c \hat{k}$
∴ Equation (i) becomes
$\begin{gathered} (a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=\lambda \\\\ \Rightarrow a+b+c=\lambda \end{gathered}$
Substituting $\lambda =a+b+c$ in equation (i), we obtain
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=a+b+c$ … (ii)
This is vector equation of the required plane
Substituting $\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$ in equation (ii)

$\begin{gathered} (x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=a+b+c \\\\ x+y+z=a+b+c \end{gathered}$

The Plane exercise 28.8 question 18

1. $13 x+14 y+11 z=0,$ This plane doesn’t satisfies the given condition.
2. The equation of the required plane is, $7 x+11 y+14 z=15$.
3. The equation of the required plane is, $7 x+11 y+4 z=33$
Hint:- By using intercept form of plane
$\frac{x}{a}+\frac{y}{b}+\frac{c}{z}=1$
Given:- The equation of the family of planes passing through the intersection of the planes
$x+2 y+3 z-4=0$ and $2 x+y-z+5=0$
Solution:- The equation of the family of planes passing through the intersection of the planes
$x+2 y+3 z-4=0$ and $2 x+y-z+5=0$
$(x+2 y+3 z-4)+k(2 x+y-z+5)=0$ , where k is some constant.
\begin{aligned} &(2 k+1) x+(k+2) y+(3-k) z=4-5 k \\\\ &\frac{x}{\left(\frac{4-5 k}{2 k+1}\right)}+\frac{y}{\left(\frac{y}{k+2}\right)}+\frac{z}{\left(\frac{4-5 k}{3-k}\right)}=1 \end{aligned}
It is given that x-intercept of the required plane is twice its z intercept.
\begin{aligned} &\left(\frac{4-5 k}{2 k+1}\right)=2\left(\frac{4-5 k}{3-k}\right) \\\\ &(4-5 k)(3-k)=(4 k+2)(4-5 k) \\\\ &(4-5 k)(3-k-4 k-2)=0 \end{aligned}
$(4-5 k)(1-5 k)=0$
Either
\begin{aligned} &4-5 k=0 \quad \text { or } \quad 1-5 k=0\\\\ &k=4 / 5 \quad \text { or }\quad k=1 / 5 \end{aligned}
When $k=4 / 5$, the equation of the plane is
\begin{aligned} &\left(2 \times \frac{4}{5}+1\right) x+\left(\frac{4}{5}+2\right) y+\left(3-\frac{4}{5}\right) z=4-5 \times \frac{4}{5} \\\\ &13 x+14 y+11 z=0 \end{aligned}
This plane does not satisfies the given condition, so this is rejected
When $k=1 / 5$ , the equation of the plane is
\begin{aligned} &\left(2 \times \frac{1}{5}+1\right) x+\left(\frac{1}{5}+2\right) y+\left(3-\frac{1}{5}\right) z=4-5 \times \frac{1}{5} \\\\ &7 x+11 y+14 z=15 \end{aligned}
Thus, the equation of the required plane is $7 x+11 y+14 z=15$.
Also, the equation of the plane passing through the point $(2,3,-1)$ and parallel to the plane $7 x+11 y+14 z=15$ is

\begin{aligned} &7(x-2)+11(y-3)+14(z+1)=0 \\\\ &7 x+11 y+4 z=33 \end{aligned}

The Plane exercise 28.8 question 19

Answer:- The answer of the given question is $x+2y+3z=4.$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+y+z=1$
$2x+3y+4z=5$
∴ Required equation of plane is $x+y-1+z+\lambda(2 x+3 y+4 z-5)=0$ for some λ
i.e.$(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z=(1+5 \lambda)$
According to question
$2\left(\frac{1+5 \lambda}{1+3 \lambda}\right)=3\left(\frac{1+5 \lambda}{1+4 \lambda}\right)$
Solving we get $\lambda =-1$
Thus the equation of required plane is
$-x-2y-3z=-4$
$x+2y+3z=4$

The RD Sharma class 12th exercise 28.8 covers the chapter 'The Plane.' There are about 19 questions in this exercise that are extremely basic and simple to solve if you have knowledge of the fundamentals of this chapter. The RD Sharma class 12th exercise 28.8 covers all the essential concepts of this chapter that are mentioned below,

• Equation of plane which is parallel to the line

• Equation of plane passing through points and parallel to the plane

• Equation of plane in scalar product form

• Equation of plane passing through line of intersection of the planes

Mentioned below are some benefits of the RD Sharma class 12 solution chapter 28 exercise 28.8:-

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