RD Sharma Class 12 Exercise 28.11 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.11 The Plane Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:53 AM IST

The RD Sharma solution books are every class 12 students’ best companion. It helps them in clarifying their doubts, helping with the homework, and sharpening their knowledge for the exams. And the syllabus of the subject mathematics is a significant threat to the class 12 students. Moreover, a concept like The Plane tends to confuse the students a lot. RD Sharma solutions In such circumstances, the RD Sharma Class 12th Exercise 28.11 reference material plays a major role.

## The Plane Excercise: 28.11

The Plane Exercise 28.11 Question 1

Answer:$\sin ^{-1} \frac{9}{\sqrt{89}}$
Hint: Use formula $\vec{r}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{k})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{k}) \text { and } \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$
Given:
Solution: Equation of line is $\vec{r}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{k})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{k})$ and the equation of plane is $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$
As we know that the angle $\theta$ between the line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
$\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Here, $a_{1}=1, b_{1}=1, c_{1}=1$
and $a_{2}=2, b_{2}=3, c_{2}=4$
The angle between them is
$\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
\begin{aligned} &=\frac{1 \times 2+1 \times 3+1 \times 4}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{2^{2}+3^{2}+4^{2}}} \\ & \end{aligned}
$=\frac{2+3+4}{\sqrt{3} \sqrt{29}} \\$
$=\frac{9}{\sqrt{87}} \\$
$\theta=\sin ^{-1} \frac{9}{\sqrt{87}}$

The Plane Exercise 28.11 Question 2

Answer:$0$
Hint: Use formula $\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}$
Given: Line is $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1}$ and plane is $2 x+y-z=4$
Solution: The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and the given plane is normal to the vector $\vec{n}=2 \hat{i}+\hat{j}-\hat{k}$
We know that, the angle $\theta$ between the line and plane is
\begin{aligned} &\sin \theta=\frac{(\hat{i}-\hat{j}+\hat{k}) \cdot(2 \hat{i}+\hat{j}-\hat{k})}{|\hat{i}-\hat{j}+\hat{k}||2 \hat{i}+\hat{j}-\hat{k}|} \\ & \end{aligned} $\left[\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right]$
$=\frac{2-1-1}{\sqrt{1+1+1} \sqrt{4+1+1}}=0 \\$
$\Rightarrow \theta=\sin ^{-1} 0 \\$
$\theta=0$

The Plane Exercise 28.11 Question 3

Answer:$\sin ^{-1}\left(\frac{23}{11 \sqrt{11}}\right)$
Hint: Use formula $\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}$
Given:$(3,-4,-2) \&(12,2,0) \text { and plane } 3 x-y+z=1$
Solution: It is given that the line passes through $(3,-4,-2) \&(12,2,0)$
So, \begin{aligned} \vec{b}=\overrightarrow{\mathrm{AB}} &=\overline{\mathrm{OB}}-\overline{\mathrm{OA}} \\ & \end{aligned}
$=12 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}-(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\$
$=9 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=9 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and given line is normal to the vector $\vec{n}=3 \hat{i}-\hat{j}+\hat{k}$
We know that, the angle $\theta$ between the line and plane is given by
\begin{aligned} \sin \theta &=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} \\ & \end{aligned}
$=\frac{(9 \hat{i}+6 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}-\hat{j}+\hat{k})}{|9 \hat{i}+6 \hat{j}+2 \hat{k} \| 3 \hat{i}-\hat{j}+\hat{k}|} \\$
$=\frac{27-6+2}{\sqrt{81+36+4} \sqrt{9+1+1}}=0 \\$
$=\frac{23}{11 \sqrt{11}} \\$
$\Rightarrow \theta =\sin ^{-1} \frac{23}{11 \sqrt{11}}$

The Plane exercise 28.11 question 4

Answer:$m = -3$
Hint: Use properties of vector
Given: Line $\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\lambda(2 \hat{\mathrm{i}}-\mathrm{m} \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ and
Plane $\vec{r} \cdot(m \hat{i}+3 \hat{j}+\hat{k})=4$
Solution: The given line is parallel to the vector $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-\mathrm{m} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and the given plane is normal to the vector $\overrightarrow{\mathrm{n}}=m \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
If the line is parallel to the plane, the normal to the plane is perpendicular to the line
\begin{aligned} &\Rightarrow \vec{b} \perp \vec{n} \\ &\Rightarrow \vec{b} \cdot \vec{n}=0 \\ &\Rightarrow(2 \hat{i}-m \hat{j}-3 \hat{k})(m \hat{i}+3 \hat{j}+\hat{k})=0 \\ &\Rightarrow 2 m-3 m-3=0 \\ &\Rightarrow-m-3=0 \\ &\Rightarrow m=-3 \end{aligned}

The Plane exercise 28.11 question 5

Answer: Therefore, the given line is parallel to given plane is showed. And distance is $\frac{7}{\sqrt{3}}$ unit
Hint: Use properties of vector and formula $d=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}$
Given: $\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \text { and } \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=7$
Solution: The given plane passes through the point with position vector $\vec{a}=2 \hat{i}+5 \hat{j}+7 \hat{k}$ and is parallel to the vector $\vec{b}=\hat{i}+3 \hat{j}+4 \hat{k}$
The given plane is $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=7$
So, the normal vector $\vec{n}=\hat{i}+\hat{j}-\hat{k} \text { and } d=7$
Now, $\vec{b} \cdot \vec{n}=(\hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})=1+3-4=0$
So, $\vec{b}$ is perpendicular to $\vec{n}$
So, the given line is parallel to the given plane.
The distance between the line and the parallel plane.
Then, d = length of perpendicular from the point $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$ to the plane $\vec{r} \cdot \vec{n}=d$
\begin{aligned} d &=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|} \\ & \end{aligned}
$=\frac{|(2 \hat{i}+5 \hat{j}+7 \hat{k})(\hat{i}+\hat{j}-\hat{k})-7|}{|\hat{i}+\hat{j}-\hat{k}|}$
\begin{aligned} &=\frac{|2+5-7-7|}{\sqrt{1+1+1}} \\ & \end{aligned}
$=\frac{7}{\sqrt{3}} \text { units }$

The Plane exercise 28.11 question 6

Answer:$\vec{r}=\lambda(\hat{i}+2 \hat{j}+3 \hat{k})$
Hint: We use properties of vector
Given:$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=3$
Solution: The requires line is perpendicular to the plane $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=3$
Therefore, it is parallel to the normal $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
Thus, the required line passes through the position vector $\overrightarrow{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}$ and parallel to the vector $\overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
So, its vector equation is
\begin{aligned} &\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \vec{n} \\ &\vec{r}=(0 \hat{i}+0 \hat{\mathrm{j}}+0 \hat{k})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ &\vec{r}=\lambda(\hat{\mathbf{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \end{aligned}

The Plane Excercise 28.11 Question 7

Answer:$7 y+4 z-5=0$
Hint: Use formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given: $(2,3,-4) \&(1,-1,3)$
Solution: We know that the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is given by
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ ……………….. (1)
So, equation of plane passing through (2,3,-4) is
$a(x-2)+b(y-3)+c(z+4)=0$ …………….…… (2)
It also passes through (1,-1,3)
$\Rightarrow-a-4 b+7 c=0 \\$
$a(1-2)+b(-1-3)+c(3+4)=0 \\$
\begin{aligned} & &\Rightarrow a+4 b-7 c=0 \end{aligned} ……………….... (3)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ ………………….. (4)
Here equation (2) is parallel to x- axis
$\frac{x}{1}=\frac{y}{0}=\frac{x}{0}$ …………………… (5)
Using (2) and (5) in equation (4) we get
\begin{aligned} &a \times 1+b \times 0+c \times 0=0 \\ &\Rightarrow a=0 \end{aligned}
Putting the value of ‘a’ in equation (3) we get
\begin{aligned} &\Rightarrow a-4 b+7 c=0 \\ & \end{aligned}
$\Rightarrow 0-4 b+7 c=0 \\$
$\Rightarrow-4 b=-7 c \\$
$\Rightarrow b=\frac{7 c}{4}$
Now, putting the value of a and b in (2) we get
\begin{aligned} &a(x-2)+b(y-3)+c(z+4)=0 \\ & \end{aligned}
$\Rightarrow 0(x-2)+\frac{7 c}{4}(y-3)+c(z+4)=0$
\begin{aligned} &\Rightarrow 0+\frac{7 c y}{4}-\frac{21 c}{4}+c z+4 c=0 \\ & \end{aligned}
$\Rightarrow 7 c y-21 c+4 c z+16 c=0$
Dividing by c we have
\begin{aligned} &7 y-21+4 z+16=0 \\ & \end{aligned}
$\Rightarrow 7 y+4 z-5=0$

The Plane Excercise 28.11 Question 8

Answer:$x-19 y-11 z=0$
Hint: Use formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given: Points $(0,0,0) \&(3,-1,2)$ and line $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}$
Solution: We know that the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is given by
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ ……………….. (1)
So, equation of plane passing through (0,0,0) is
$a(x-0)+b(y-0)+c(z-0)=0$
\begin{aligned} &\\ &a x+b y+c z=0 \end{aligned} …………………. (2)
It also passes through $(3,-1,2)$
So, equation (2) must satisfy the point $(3,-1,2)$
$3 a-b+2 c=0$ ………………… (3)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …………........ (4)
Here, the plane is parallel to line
$\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7} \\$
So, $a \times 1+b \times(-4)+c \times 7=0 \\$
\begin{aligned} & &\Rightarrow a-4 b+7 c=0 \end{aligned} …………….… (5)
Solving equation (3) and (5) by cross multiplication we have,
\begin{aligned} &\frac{\mathrm{a}}{-1 \times 7-(-4) \times 2}=\frac{\mathrm{b}}{1 \times 2-3 \times 7}=\frac{\mathrm{c}}{3 \times(-4)-1 \times(-1)} \\ & \end{aligned}
$\Rightarrow \frac{\mathrm{a}}{-7+8}=\frac{\mathrm{b}}{2-21}=\frac{\mathrm{c}}{-12+1}$
\begin{aligned} &\Rightarrow \frac{a}{1}=\frac{b}{-19}=\frac{c}{-11}=k \\ & \end{aligned}
$a=k, b=-19 k, c=-11 k$
Putting the value in equation (2) we get
\begin{aligned} &a x+b y+c z=0 \\\\ & \end{aligned}
$k x-19 k y-11 k z=0$
Dividing by k we have
$x-19 y-11 z=0$
The required equation is $x-19 y-11 z=0$

The Plane Excercise 28.11 Question 9

Answer:$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k(-3 \hat{i}+5 \hat{j}+4 \hat{k})$
or
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
Hint: Use formula $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$
Given: Point $\left ( 1,2,3 \right )$ and planes
\begin{aligned} &\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \\ & \end{aligned}
$\vec{r} \cdot(3 \hat{i}+\hat{j}+2 \hat{k})=6$
Solution: We know that, the equation of line passing through $\left ( 1,2,3 \right )$ is given by
$\frac{x-1}{a_{1}}=\frac{y-2}{b_{1}}=\frac{z-3}{c_{1}}$ ……………….. (1)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
If $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ ………………… (2)
Here, line (1) is parallel to plane
$x-y+2 z=5 \\$
So, $a \times 1+b \times(-1)+c \times 2=0 \\$
$\begin{gathered} \Rightarrow a-b+2 c=0 \end{gathered}$ …………………. (3)
Also, line (1) is parallel to plane,
$3 x+y+z=6$
So, $a \times 3+b \times(1)+c \times 1=0 \\$
\begin{aligned} & &3 a+b+c=0 \end{aligned} ………………….. (4)
Solving equation (3) and (4) by cross multiplication
We have,
\begin{aligned} &\frac{a}{-1 \times 1-1 \times 2}=\frac{b}{3 \times 2-1 \times 1}=\frac{c}{1 \times 1-3 \times(-1)} \\ & \end{aligned}
$\Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3} \\$
$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k \\$
$a=-3 k, b=5 k, c=4 k$
Putting the value in equation (1) we have
$\frac{x-1}{-3 k}=\frac{y-2}{5 k}=\frac{z-3}{4 k}$
Multiplying by k we have
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
The required equation is
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}\\$
or
\begin{aligned} & & &\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k(-3 \hat{i}+5 \hat{j}+4 \hat{k}) \end{aligned}

The Plane Exercise 28.11 Question 10

Answer: The line of section is parallel to plane
Hint: Use properties of plane
Given:
\begin{aligned} &5 x+2 y-4 z+2=0,2 x+8 y+2 z-1=0 \\\\ &\& 4 x-2 y-5 z-2=0 \end{aligned}
Solution: Let $a_{1}, b_{1} \& c_{1}$ be the direction ratios of line $5 x+2 y-4 z+2=0,2 x+8 y+2 z-1=0$
As we know, that if two planes are perpendicular with direction ratios as $a_{1}, b_{1} \& c_{1}$ and $a_{2}, b_{2} \& c_{2}$ then $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
Since line lies in both the planes, so it is perpendicular to both planes
\begin{aligned} &5 a_{1}+2 b_{1}-4 c_{1}=0 \\ & \end{aligned} …………….. (1)
$2 a_{1}+8 b_{1}+2 c_{1}=0$ …………….. (2)
Solving equation (1) and (2) by cross multiplication
We have,
\begin{aligned} &\frac{a_{1}}{2 \times 2-(-4) \times 8}=\frac{b_{1}}{2 \times(-4)-5 \times 2}=\frac{c_{1}}{5 \times 8-2 \times 2} \\ & \end{aligned}
$\Rightarrow \frac{a_{1}}{4+32}=\frac{b_{1}}{-8-10}=\frac{c_{1}}{40-4}$
\begin{aligned} &\Rightarrow \frac{a_{1}}{36}=\frac{b_{1}}{-18}=\frac{c_{1}}{36}=k \\ & \end{aligned}
$\Rightarrow \frac{a_{1}}{2}=\frac{b_{1}}{-1}=\frac{c_{1}}{2}=k \\$
$a=2 k, b=-k, c=2 k$
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ ………….. (3)
Here, line with direction ratios is parallel to plane
\begin{aligned} &4 x-2 y-5 z-2=0 \\ & \end{aligned}
$\Rightarrow 2 \times 4+(-1) \times(-2)+2 \times(-5)=0 \\$
$\Rightarrow 8+2-10=0$
Therefore, the line of section is parallel to the plane.

The Plane Exercise 28.11 Question 11

$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mathrm{k}((2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}))$
Hint: Use properties of plane
Given: Point $(1,-1,2)$ and plane $2 x-y+3 z-5=0$
Solution: Equation of line passing through $\vec{a}$ and $\vec{b}$ is given by $\vec{r}=\vec{a}+k \vec{b}$ ……………. (1)
Given that the line passes through $(1,-1,2) \text { is } \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+k \vec{b}$ …………….. (2)
Since, line (1) is perpendicular to the plane $2 x-y+3 z-5=0$
So, normal to plane is parallel to the line.
In vector form,
$\overrightarrow{\mathrm{b}}=\mathrm{m}(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ as, is any scalar.
Thus, the equation of required line,
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mathrm{k}((2 \hat{\mathbf{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}))$

The Plane Exercise 28.11 Question 12

Answer: $12 x+15 y-14 z-68=0$
Hint: Use formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given: Points $(2,2,-1) \&(3,4,2)$ ratios $7, 0, 6$
Solution: We know that the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ ……………….. (1)
So, equation of plane passing through $(2,2,-1)$ is
$a(x-2)+b(y-2)+c(z+1)=0$ ………………… (2)
It also passes through $(3,4,2)$
So, equation (2) passes through $(3,4,2)$
\begin{aligned} &a(3-2)+b(4-2)+c(2+1)=0 \\ & \end{aligned}
$\Rightarrow a+2 b+3 c=0$ …………………. (3)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …………… (4)
Here, the plane (2) is parallel to line having direction ratios 7, 0 ,6
So, \begin{aligned} &a \times 7+b \times 0+c \times 6=0 \\ & \end{aligned}
$\Rightarrow 7 a+6 c=0 \\$
$\Rightarrow a=\frac{-6 c}{7}$ ………………… (5)
Putting the value of a in equation (3) we get
\begin{aligned} &a+2 b+3 c=0 \\ & \end{aligned}
$\Rightarrow \frac{-6 c}{7}+2 b+3 c=0 \\$
$\Rightarrow 14 b+15 c=0 \\$
$\Rightarrow b=\frac{-15 c}{14}$
Putting the value of a and b in equation (2) we get
\begin{aligned} &a(x-2)+b(y-2)+c(z+1)=0 \\ & \end{aligned}
$\Rightarrow \frac{-6 c}{7}(x-2)+\frac{-15 c}{14}(y-2)+c(z+1)=0 \\$
$\Rightarrow \frac{-6 x c}{7}+\frac{12 c}{7}-\frac{15 c y}{14}+\frac{30 c}{14}+c z+c=0$
Multiplying by $\frac{14}{c}$ we have
\begin{aligned} &\Rightarrow-12 x+24-15 y+30+14 z+14=0 \\ \end{aligned}
$\Rightarrow-12 x-15 y+14 z+68=0 \\$
$\Rightarrow 12 x+15 y-14 z-68=0$
Equation of required plane is $12 x+15 y-14 z-68=0$

The Plane exercise 28.11 question 13

Answer:$\sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$
Hint: Use formula $\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given: Line is $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$ and plane is $3 x+4 y+z+5=0$
Solution: We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
If $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
and the angle between them is given by
$\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$ …………….. (1)
Now, given equation by line is
$\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$
So, $a_{1}=3, b_{1}=-1, c_{1}=2$
Equation of plane is $3 x+4 y+z+5=0$
So, \begin{aligned} a_{2}=& 3, b_{2}=4, c_{2}=1 \& d_{2}=-5 \\ \end{aligned}
$\sin \theta=\frac{3 \times 3+(-1) \times 4+2 \times 1}{\sqrt{3^{2}+(-1)^{2}+2^{2}} \sqrt{3^{2}+4^{2}+1^{2}}} \\$
$\sin \theta =\frac{9-4+2}{\sqrt{9+1+4} \sqrt{9+16+1}}$
\begin{aligned} &=\frac{7}{\sqrt{14} \sqrt{26}} \times \frac{\sqrt{7}}{\sqrt{7}} \\ & \end{aligned}
$=\frac{7 \sqrt{7}}{7 \sqrt{52}}=\frac{\sqrt{7}}{\sqrt{52}}$
$\sin \theta=\frac{\sqrt{7}}{\sqrt{52}} \Rightarrow \theta=\sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$
The angle between the plane and the line is $\theta=\sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)$

The Plane exercise 28.11 question 14

Answer: Therefore the equation of the plane is $9 x-8 y+7 z-21=0$ and distance is $\mathrm{D}=\frac{13}{\sqrt{194}}$
Hint: Use formula $a x+b y+c z+d=0$ and $D=\frac{a x_{1}+b y_{1}+c z_{1}+d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}$
Given: Planes are $x-2 y+z=1 \& 2 x+y+z=8$ and point $\left ( 1,1,1 \right )$ ratios proportional is $1, 2, 1.$
Solution: We know that equation of plane passing through the intersection of 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
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
So, equation of plane passing through the intersection of planes $x-2 y+z-1=0 \& 2 x+y+z-8=0$ is
\begin{aligned} &\Rightarrow(x-2 y+z-1)+k(2 x+y+z-8)=0 \\ & \end{aligned} ………………. (1)
$\Rightarrow x(1+2 k)+y(-2+k)+z(1+k)+(-1-8 k)=0$
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$
Is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \; \text { if }\; a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
Given the plane is parallel to line with direction ratios 1, 2, 1
\begin{aligned} &1 \times(1+2 k)+2 \times(-2+k)+1 \times(1+k)=0 \\ & \end{aligned}
$\Rightarrow 1+2 k-4+2 k+1+k=0 \\$
$\Rightarrow k=\frac{2}{5}$
Putting the value of k in equation (1) we get
\begin{aligned} &\Rightarrow(x-2 y+z-1)+\frac{2}{5}(2 x+y+z-8)=0 \\ & \end{aligned}
$\Rightarrow x+\frac{4}{5} x-2 y+\frac{2}{5} y+z+\frac{2}{5} z-1-\frac{16}{5}=0 \\$
$\Rightarrow \frac{9 x}{5}-\frac{8 y}{5}+\frac{7 z}{5}-\frac{21}{5}=0 \\$
$\Rightarrow 9 x-8 y+7 z-21=0$
We know that the distance (D) of point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ from plane $a x+b y+c x-d=0$ is given by
$D=\frac{a_{1} x+b_{1} y+c_{1} z+d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}$
So, distance of Point $\left ( 1,1,1 \right )$ from plane is
\begin{aligned} D &=\frac{9 \times 1+(-8) \times 1+7 \times 1-21}{\sqrt{9^{2}+(-8)^{2}+7^{2}}} \\ \end{aligned}
$=\frac{9-8+7-21}{\sqrt{81+64+49}} \\$
$=\frac{-13}{\sqrt{194}}$
Taking the mod value we have $D=\frac{13}{\sqrt{194}} \text { unit }$

The Plane exercise 28.11 question 15
Answer: Therefore, the line $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$ and plane $\vec{r} \cdot \vec{n}=d$ are parallel. And line

$\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ is parallel to $\vec{r} \cdot(2 \hat{j}+\hat{k})=3 .$. The distance between lines and plane is $\frac{1}{\sqrt{5}}$ units.
Hint: Use formula $\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}$
Given:$\vec{r}=\hat{i}+\hat{j}+\lambda(3 \hat{i}-\hat{j}+2 \hat{k}) \text { and } \vec{r} \cdot(2 \hat{j}+\hat{k})=3$
Solution: We know that line $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$ and plane $\vec{r} \cdot \vec{n}=d$ is parallel
if $\vec{b} \cdot \vec{n}=0$ ……………… (1)
Given, equation of line $\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ and equation of plane is $\vec{r} \cdot(2 \hat{j}+\hat{k})=3$
So, $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{n}}=2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Now, $\overrightarrow{\mathrm{b}} \cdot \vec{n}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=-2+2=0$
So, the line and plane are parallel.
We know that, the distance ‘D’ of plane is $\vec{r} \cdot \vec{n}=d$ from a point $\vec{a}$ is given by
\begin{aligned} &\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}} \\ & \end{aligned}
$\overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
\begin{aligned} D &=\frac{(\hat{i}+\hat{j})(2 \hat{j}+\hat{k})-3}{\sqrt{2^{2}+1^{2}}} \\ & \end{aligned}
$=\frac{2-3}{\sqrt{4+1}}=\frac{-1}{\sqrt{5}}$
We take the mod value
$D=\frac{1}{\sqrt{5}}$

The Plane Exercise 28.11 Question 16

Answer:$\frac{1}{\sqrt{6}}$ units
Hint:$\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}$
Given:$\vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1 \text { and } \vec{r}=(-\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}+4 \hat{k})$
Solution: We know that line $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{kb}$ and plane $\vec{r} \cdot \vec{n}=\mathrm{d}$ is parallel if $\vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1$
Given, equation of line $\vec{r}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{k})+k(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{k})$ and equation of plane is $\vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1$
So, $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
Now,
So, the line and plane are parallel
we know that the Distance ‘D’ of a plane $\vec{r} \cdot \vec{n}=d$ from a point $\vec{a}$ is given by
$\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}, \overrightarrow{\mathrm{a}}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$$\mathrm{D}=\frac{1}{\sqrt{6}} \text { units }$
\begin{aligned} \Rightarrow D &=\frac{(-\hat{i}+\hat{j}+\hat{k})(\hat{i}+2 \hat{j}-\hat{k})-1}{\sqrt{(-1)^{2}+2^{2}+1^{2}}} \\ & \end{aligned}
$=\frac{-1+2-1-1}{\sqrt{1+4+1}}=\frac{-1}{\sqrt{6}}$
We take the mod value,
So, $\mathrm{D}=\frac{1}{\sqrt{6}} \text { units }$

The Plane Exercise 28.11 Question 17

Answer:$x-20 y+27 z-14=0$
Hint: Use Properties of plane
Given:$3 x-4 y+5 z=10 \text { and } 2 x+2 y-3 z=4 \text { and line } x=2 y=3 z$
Solution: We know that equation of plane passing through the intersection of plane $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
So, equation of plane passing through the intersection of planes
\begin{aligned} &3 x-4 y+5 z=10 \text { and } 2 x+2 y-3 z=4 \text { is } \\ \end{aligned}
$(3 x-4 y+5 z-10)+k(2 x+2 y-3 z-4)=0 \\$ ………………. (1)
$\Rightarrow x(3+2 k)+y(-4+2 k)+z(5-3 k)+(-10-4 k)=0$
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \text { if } a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
Given the plane is parallel to line $x=2 y=3 z$
Or, $\frac{x}{6}=\frac{y}{3}=\frac{z}{2}$ (dividing the equation by 6)
\begin{aligned} &6 \times(3+2 k)+3 \times(-4+2 k)+2 \times(5-3 k)=0 \\ & \end{aligned}
$\Rightarrow 18+12 k-12+6 k+10-6 k=0 \\$
$\Rightarrow k=\frac{-16}{12}=\frac{-4}{3}$
Putting the value of k in equation (1) we get
\begin{aligned} &(3 x-4 y+5 z-10)-\frac{4}{3}(2 x+2 y-3 z-4)=0 \\ & \end{aligned}
$\Rightarrow 3 x-\frac{8}{3} x-4 y-\frac{8}{3} y+5 z+4 z-10+\frac{16}{3}=0$
\begin{aligned} &\Rightarrow \frac{x}{3}-\frac{20 y}{3}+9 z-\frac{14}{3}=0 \\ & \end{aligned}
$\Rightarrow x-20 y+27 z-14=0$
The required equation is $x-20 y+27 z-14=0$

The Plane Exercise 28.11 Question 18

Answer: Therefore, required equation of plane is $\vec{r} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=11 \text { and }-9 x+8 y-z=11$ and distance of plane is $\sqrt{146} \: units$
Hint: Use properties of plane
Given:\begin{aligned} &\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & \end{aligned}
$\vec{r}=(\hat{i}-3 \hat{j}+5 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$
Solution: The plane passes through the point $\left ( 1,2,-4 \right )$
A vector in a direction perpendicular to $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+m(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})+\mathrm{n}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \text { is } \overrightarrow{\mathrm{n}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
$\Rightarrow \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{array}\right|=-9 \hat{i}+8 \hat{\mathrm{j}}-\hat{k}$
Equation of the plane is $(\vec{r}-\vec{a}) \cdot \vec{n}=0$
\begin{aligned} &\{\vec{r}-(\hat{i}+2 \hat{j}-4 \hat{k})\} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=0 \\ & \end{aligned}
$\Rightarrow \vec{r} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=11$
Substitution $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ , we get the Cartesian from as $-9 x+8 y-z=11$
The distance of the point $(9,8,-10)$ from the plane
\begin{aligned} &=\frac{-9 \times 9+8 \times(-8)-1 \times(-10)-11}{\sqrt{-9^{2}+8^{2}+-1^{2}}} \\ & \end{aligned}
$=\frac{-81-64+10-11}{\sqrt{81+64+1}} \\$
$=\left|\frac{-146}{\sqrt{146}}\right|=\sqrt{146} \text { units }$

The Plane Excercise 28.11 Question 19

Answer: Therefore, required equation of plane is $8 x-13 y+15 z+13=0$
Hint: Use formula $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
Given:$\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}$
Solution: We know that the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ is given by
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ ……………… (1)
So, equation of plane passing through $\left ( 3,4,1 \right )$ is
$a(x-3)+b(y-4)+c(z-1)=0$ ……………… (2)
It also passes through $\left ( 0,1,0 \right )$
So, equation (2) must satisfy the point $\left ( 0,1,0 \right )$
\begin{aligned} &a(0-3)+b(1-4)+c(0-1)=0 \\ \end{aligned}
$\Rightarrow-3 a-3 b-c=0 \\$
$\Rightarrow 3 a+3 b+c=0$ ……………… (3)
We know that line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ if
\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ & \end{aligned} ……………… (4)
So, $a \times 2+b \times 7+c \times 5=0 \\$
$\Rightarrow 2 a+7 b+5 c=0$ ………………. (5)
Solving equation (3) and (5) by cross a multiplication, we have
\begin{aligned} &\frac{\mathrm{a}}{3 \times 5-(7) \times 1}=\frac{\mathrm{b}}{2 \times 1-3 \times 5}=\frac{\mathrm{c}}{3 \times 7-2 \times 3} \\ & \end{aligned}
$\Rightarrow \frac{\mathrm{a}}{15-7}=\frac{\mathrm{b}}{2-15}=\frac{\mathrm{c}}{21-6}$
\begin{aligned} &\Rightarrow \frac{a}{8}=\frac{b}{-13}=\frac{c}{15}=k(s a y) \\ \end{aligned}
$a=8 k, b=-13 k, c=15 k$
Putting the value in equation (2) we get
\begin{aligned} &8 k(x-3)-13 k(y-4)+15 k(z-1)=0 \\ & \end{aligned}
$8 k x-24 k-13 k y+52 k+15 k z-15 k=0$
Dividing by k we have
$8 x-13 y+15 z+13=0$
Equation of required plane is $8 x-13 y+15 z+13=0$

The Plane Excercise 28.11 Question 20

Answer: Coordinates of intersection is $(2,-1,2), \sin ^{-1}\left(\frac{1}{\sqrt{27}}\right)$
Hint: Use formula $\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} \text { and } x-y+z-5=0$
Solution: Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=r$
$\Rightarrow x=3 r+2, y=4 r-1, z=2 r+2$
Substituting in the equation of the plane $x-y+z-5=0$ , We get,
\begin{aligned} &(3 r+2)-(4 r-1)+(2 r+2)-5=0 \\ & \end{aligned}
$\Rightarrow 3 r+2-4 r+1+2 r+2-5=0 \\$
$\Rightarrow r=0$
\begin{aligned} &x=3 \times 0+2 \Rightarrow x=2 \\ \end{aligned}
$y=4 \times 0-1 \Rightarrow y=-1 \\$
$z=2 \times 0+2 \Rightarrow z=2$
Hence, the coordinates of intersection is $(2,-1,2)$
Direction ratios of the line are $3, 4, 2$
Direction ratios of the line perpendicular to the plane are $1, -1, 1$
\begin{aligned} &\sin \theta=\frac{3 \times 1+(-1) \times 4+2 \times 1}{\sqrt{3^{2}+4^{2}+2^{2}} \sqrt{1^{2}+(-1)^{2}+1^{2}}} \\ & \end{aligned}
$\sin \theta=\frac{3-4+2}{\sqrt{9+16+4} \sqrt{1+1+1}}$
\begin{aligned} &=\frac{1}{\sqrt{29} \sqrt{3}} \\ \end{aligned}
$=\frac{1}{\sqrt{87}}$
$\theta=\sin ^{-1}\left(\frac{1}{\sqrt{87}}\right)$
The angle between the plane and line is $\sin ^{-1}\left(\frac{1}{\sqrt{87}}\right)$

The Plane Excercise 28.11 Question 21

Answer:$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
Hint:$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{kb}$
Given:$\vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9=0 \text { and }(1,2,3)$
Solution: We know that equation of line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r}=\vec{a}+k \vec{b}$ …………. (1)
Given that, the line is passing through $\left ( 1,2,3 \right )$
So, $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$
It is given that line is perpendicular to plane $\vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9=0$
So, normal to plane $\left ( \vec{n} \right )$ is parallel to $\vec{b}$
So, Let $\vec{b}=\vec{n}=1(\hat{i}+2 \hat{j}-5 \hat{k})$
Putting $\vec{a} \& \vec{b}$ in (1), equation of line is
\begin{aligned} &\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k\{1(\hat{i}+2 \hat{j}-5 \hat{k})\} \\ & \end{aligned}
$\Rightarrow \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+m(\hat{i}+2 \hat{j}-5 \hat{k})$

The Plane Exercise 28.11 Question 22

Answer:$\sin ^{-1}\left(\frac{8}{21}\right)$
Hint: Use formula $\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Given:$\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \text { and } 10 x+2 y-11 z=3$
Solution: Direction ratios of the line
$\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \text { are }(2,3,6)$
Direction ratio of a line perpendicular to the plane $10 x+2 y-11 z=3 \text { are }(10,2,-11)$
As we know that the angle $\theta$ between the line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and plane
$a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
\begin{aligned} &\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ & \end{aligned}
$\sin \theta=\frac{2 \times 10+3 \times 2+6 \times(-11)}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+(-11)^{2}}}$
\begin{aligned} \sin \theta &=\frac{-40}{\sqrt{49} \sqrt{225}} \\ & \end{aligned}
$=\frac{-40}{7 \times 15}=\frac{-8}{21}$
$\begin{gathered} \sin \theta=\frac{-8}{21} \\ \end{gathered}$
$\theta=\sin ^{-1}\left(\frac{8}{21}\right)$

The Plane Exercise 28.11 Question 23

Answer:$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$
Hint: Use the properties of plane
Given:$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=5 \text { and } \overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=6$
Solution: We know that, the equation of line passing through $\left ( 1,2,3 \right )$ is given by
$\frac{x-1}{a_{1}}=\frac{y-2}{b_{1}}=\frac{z-3}{c_{1}}$ ………….. (1)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …………… (2)
Here line (1) is parallel to plane
$\begin{gathered} x-2 y+z=5 \\ \end{gathered}$
So, $a \times 1+b \times(-1)+c \times 2=0 \\$
$\Rightarrow a-b+2 c=0$ …………... (3)
Also, line (1) parallel to plane
$\begin{gathered} 3 x+y+z=6 \\ \end{gathered}$
So, $a \times 3+b \times(1)+c \times 1=0 \\$
$\Rightarrow 3 a+b+c=0$ …………… (4)
Solving equation (3) and (4) by cross multiplication we have,
\begin{aligned} &\frac{a}{-1 \times 1-(1) \times 2}=\frac{b}{3 \times 2-1 \times 1}=\frac{c}{1 \times 1-3 \times(-1)} \\ & \end{aligned}
$\Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3} \\$
$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k(\text { say })$
$a=-3 k, b=5 k \& c=4 k$
Putting the value in equation (1) we get
$\frac{x-1}{-3 k}=\frac{y-2}{5 k}=\frac{z-3}{4 k}$
Multiplying by k we have
$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
The required equation is
$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})$

The Plane Exercise 28.11 Question 24

Answer:$-2$
Hint: Use formula $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$
Given:$\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4} \text { and } 3 x-y-2 z=7$
Solution: Here given midline
$\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{4}$ is perpendicular to plane $3 x-y-2 z=7$
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is perpendicular to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ if
$\frac{a_{1}}{a_{2}}+\frac{b_{1}}{b_{2}}+\frac{c_{1}}{c_{2}}=0$
So, normal vector of plane is parallel to line.
So, direction ratios of normal to plane are proportional to the direction of line.
Here, $\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
By cross multiplying the last two we have
$-2 \lambda=4 \Rightarrow \lambda=-2$

The Plane Excercise 28.11 Question 25

Answer:$x+2 y+3 z-3=0$
Hint: Use formula $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$
Given: Points $(-1,2,0) \&(2,2,-1)$ and line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$
Solution: We know that, the equation of plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ Is given by
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$ …………. (1)
So, equation of plane passing through $\left ( -1,2,0 \right )$ is
$a(x+1)+b(y-2)+c(z-0)=0$ …………. (2)
It also passes through $\left ( 2,2,-1 \right )$
So, equation (2) must satisfy the point $\left ( 2,2,-1 \right )$
\begin{aligned} &a(2+1)+b(2-2)+c(-1-0)=0 \\ & \end{aligned}
$\Rightarrow 3 a-c=0$ ……….. (3)
We know that line
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
If $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ ………… (4)
Here the plane is parallel to line
$\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$
So, $a \times 1+b \times 1+c \times-1=0 \\$
$\Rightarrow a+b-c=0$ ………… (5)
Solving equation (3) and (5) by cross multiplication we have,
\begin{aligned} &\frac{a}{0-(-1) \times 1}=\frac{b}{1 \times(-1)+3 \times 1}=\frac{c}{3 \times 1-0} \\ & \end{aligned}
$\Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{3}=k(\text { say })$
$a=k, b=2 k, c=3 k$
Putting the value in equation (2) we get
\begin{aligned} &k(x+1)+2 k(y-2)+3 k(z-0)=0 \\ & \end{aligned}
$k x+k+2 k y-4 k+3 k z=0$
Dividing by k we have,
$x+2 y+3 z-3=0$
The required equation is $x+2 y+3 z-3=0$

The Plane Excercise 28.11 Question 26

Answer: Required equation of plane is $x-y+z=0$ and angle $\sin \theta=\frac{-1}{\sqrt{3}}$
Hint: Use formula $\sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}$
Given: Point $(2,1,-1) \text { and } 2 x+y-z=3 \text { and } x+2 y+z=2$
Solution:
Line of intersection $\mathrm{n}_{1} \times \mathrm{n}_{2}$
$\vec{A} \vec{R} \cdot\left[n_{1} \times n_{2}\right]=0$
Scalar triple product, $\left[\begin{array}{lll} \overrightarrow{A R} & n_{1} & n_{2} \end{array}\right]$
$\overrightarrow{A R}=$ Position Vector of $\overrightarrow{ R}$ – Position Vector of $\overrightarrow{ A}$
\begin{aligned} &=(x-2) \hat{i}+(y-1) \hat{j}+(z+1) \hat{k} \\ & \end{aligned}
$\left|\begin{array}{ccc} x-2 & y-1 & z+1 \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{array}\right|=0$
\begin{aligned} &(x-2)(1+2)-(y-1)(2+1)+(z+1)(4-1)=0 \\ & \end{aligned}
$x-2-y+1+z+1=0$
$x-y+z=0$ is required equation of plane
Now, the angle between the plane and the y-axis is $\sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}$
\begin{aligned} &\sin \theta=\frac{-1}{\sqrt{1+1+1} \sqrt{0+1+0}} \\ & \end{aligned}
$\sin \theta=\frac{-1}{\sqrt{3}} \\$ $\left[\sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}\right]$
$\theta=\sin ^{-1} \frac{-1}{\sqrt{3}}$

The chapter 28 portion for class 12 mathematics contains fifteen exercises. Whereas, in the eleventh exercise, 28.11, most of the concepts revolve around finding the angle between the line and plane, angle between the point and the plane, finding the vector and cartesian equations of the line passing through the points and parallel to the plane. All the solutions for the 26 questions present in this exercise can be found in the RD Sharma Class 12 Chapter 28 Exercise 28.11 solution book. Moreover, the questions are given only under the Level 1 section, reducing the complexity.

The RD Sharma Class 12th Exercise 28.11 consists of multiple questions for practice apart from the ones given in the textbook. With a bit of practice in these concepts, the students will be able to score higher than their previous performance. This eventually elevates their confidence to face the public exam. As the RD Sharma books follow the NCERT pattern, it becomes easier for the CBSE school students to adapt to its methods.

The benefits of the RD Sharma books are ineffable as most of the students have benefitted and grown to great heights based on their 12th public exams. Any confusion regarding The Plane concept will be cleared once the students refer to the Class 12 RD Sharma Chapter 28 Exercise 28.11 Solution book. The answers are given by the teachers and experts who have a piece of in-depth knowledge in this particular domain. Therefore, the students tend to learn the concepts easier without any extra effort.

The RD Sharma Class 12 Solutions the Plane Ex 28.11 reference book consists of various methods to solve a sum. The students can try and figure out the ones that they find easy to adapt. In such a way, everyone would have their own easy way to arrive at a solution, even for a challenging sum. In this way, the RD Sharma Class 12th Exercise 28.11 book helps them a lot.

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## RD Sharma Chapter wise Solutions

1. Which solution book is the must for the students to be clear of the concepts in class 12, mathematics, chapter 28?

The RD Sharma Class 12th Exercise 28.11 reference book provides the optimal solution for Chapter 28 in mathematics.

2. How many questions are present in chapter 28, ex 28.11 in mathematics?

There are around nineteen questions given in the textbook for exercise 28.11. The RD Sharma Class 12th Exercise 28.11 given the accurate solutions for these questions.

3. Is the concept of the plane in mathematics easy to solve?

Any mathematical concept can be solved easily with a good amount of practice referring to the proper solutions book, and the concept of the plane is no exception to it.

4. Where can the RD Sharma books be found online?

The Career 360 website contains the best set of RD Sharma Solutions books for every grade and subject.

5. Is it easy for the CBSE students to use the RD Sharma solution books?

The RD Sharma solution books follow the NCERT pattern, which makes it convenient for the CBSE students to follow it.

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