RD Sharma Class 12 Exercise 28.15 The Plane Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 28.15 The Plane Solutions Maths - Download PDF Free Online

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

RD Sharma Books are the teacher's favorite, and if a student understands and learns from the concepts from this book. The RD Sharma Class 12th Exercise 28.15 solutions are crafted by subject experts who provide easy methods to solve every question. This particular exercise has 15 questions to clarify the concept in a better way.

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

This chapter, 'The Plane,' discusses 2D surfaces having infinite dimensions with no thickness such that when two lines join them, both of them lie on the surface entirely and find the shortest distance between them.

## The Plane Excercise: 28.15

The Plane exercise 28.15 question 1

Answer: $\left(-\frac{6}{61}, \frac{-8}{61}, \frac{12}{61}\right)$
Hint:
Suppose point $P(0,0,0)$, Let$P M \perp \text { Plane }$. Direction of $P M \div 3,4,-6$
Equation of line$=\frac{x-x_{1}}{\alpha}=\frac{y-y_{1}}{\beta}=\frac{z-z_{1}}{\gamma}$
$P(0,0,0)$

Given:

Find the image of the point $(0,0,b)$ in the plane $3x+4y-6z+1=0$
Solution:
\begin{aligned} &3 x+4 y-6 z+1=0 \\\\ &\text { Let } x_{1}=0=y_{1}=31 \end{aligned}
Equation of line $=\frac{x-0}{3}=\frac{y-0}{4}=\frac{z-0}{-6}$
\begin{aligned} &\text { Say } \lambda, \\ &x=3 \lambda \Rightarrow x=3 \times\left(\frac{-1}{61}\right)=\frac{-3}{61} \\ &y=4 \lambda \Rightarrow y=4 \times\left(\frac{-1}{61}\right)=\frac{-4}{61} \\ &z=-6 \lambda \Rightarrow y=-6 \times\left(\frac{-1}{61}\right)=\frac{6}{61} \end{aligned}
\begin{aligned} &\therefore x=\frac{-3}{61}, y=\frac{-4}{61}, z=\frac{6}{61} \\\\ &\frac{A}{z}=\frac{-3}{61}, \frac{B}{z}=\frac{-4}{61}, \frac{C}{z}=\frac{6}{61} \end{aligned}

The Plane exercise 28.15 question 2

Answer: $\left(\frac{73}{25},-\frac{6}{5}, \frac{39}{25}\right)$
Hint:
Suppose point $P(1,2,-1)$, Let $P M \perp \text { Plane. }$
Direction $= (3,-5,4)$
$P(1,2,-1)$

Given:

Find the reflection of the point $(1,2,-1)$ in the plane $3x-5y+4z=5$

Solution:
Direction of line $=(3,-5,4)$$=(3,-5,4)$$=(3,-5,4)$$= (3,-5,4)$
Equation of line $\frac{x-1}{3}=\frac{y-2}{5}=\frac{3+1}{4}=x$

$x=3\lambda +1,y=-5x+2,z=4\lambda -1$

Put in $(3x-5y+4z=5)$

$3(3\lambda +1)-5(-5\lambda +2)+4(4\lambda -1)=5$
\begin{aligned} &9 \lambda+3+25 \lambda-10+16 \lambda-4=5 \\\\ &50 \lambda=16 \\\\ &\lambda=\frac{16}{50}=\frac{8}{25} \end{aligned}

So that
\begin{aligned} &x=3\left(\frac{8}{25}\right)+1=\frac{73}{25} \\\\ &y=-5\left(\frac{8}{25}\right)+2=\frac{6}{5} \\\\ &z=4\left(\frac{8}{25}\right)-1=\frac{39}{25} \end{aligned}
coordinate $\left(\frac{73}{25}, \frac{-6}{5}, \frac{39}{25}\right)$ Ans.

The Plane exercise 28.15 question 3

Answer: $(1,6,0), 2 \sqrt{6}$
Hint:
Suppose $\frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}=\lambda$
$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}\left(y_{2}-y_{1}\right)^{2}\left(z_{2}-z_{1}\right)^{2}}$
Given:
Find the coordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ to the line
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ hence or otherwise deduce the length of the perpendicular.
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}$
$P(5,4,2)$

Solution:
\begin{aligned} &x=2 \lambda+1 \\\\ &y=3 \lambda+3 \\\\ &z=-\lambda+1 \end{aligned}
Direction ratio of $P$ as $=(2 \lambda-1-5,3 \lambda+z-4,-\lambda+1-2)$
\begin{aligned} &=(2 \lambda-6,3 \lambda-1,-\lambda-1) \\\ &a_{1} \quad b_{1} \quad c_{1} \\\\ &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \end{aligned}
\begin{aligned} &=2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0 \\\\ &=4 \lambda-12+9 \lambda-3+\lambda+1=0 \\\\ &=14 \lambda-14=0 \\\\ &\lambda=1 \end{aligned}
Putting the value of x in eq
\begin{aligned} &x=2 \lambda+1, \\\\ &y=3 \lambda+3, \\\\ &z=-\lambda+1 \end{aligned}
\begin{aligned} &x=3 \\ &y=6 \\ &z=0 \end{aligned}
co-ordinates $(1,6,0)$
\begin{aligned} &P Q=\sqrt{(5-1)^{2}+(4-8)^{2}+(2-0)^{2}} \\\\ &=\sqrt{16+14+4} \\\\ &=\sqrt{24} \\\\ &=2 \sqrt{6} \; \mathrm{Ans} \end{aligned}

The Plane exercise 28.15 question 4

Answer: $(1,2,1), 2 i+\frac{3}{2} \hat{\jmath}+\frac{3}{2} \hat{k}$
$\vec{r}=3 i+\hat{\jmath}+2 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k})$
Hint:
Point $(3,1,2)$
Suppose $\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
Given:
Find the image of the point with position vector $3 i+2 \hat{\jmath}+\hat{k}$ in the plane $\vec{r} \cdot(2 \hat{\imath}+\hat{j}+\hat{k})=\mathrm{Y} .$. Also find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3 i+2 \hat{\jmath}+\hat{k}$ $P(3,1,2)$

$Q(1,2,1)$
Solution:
\begin{aligned} &\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\\\ &2 x-y+z=4 \\\\ &\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1} \end{aligned}
\begin{aligned} &=\frac{-2(6-1+2-4)}{(4+1+1)} \\\\ &=\frac{-2(3)}{6} \\\\ &=-1 \end{aligned}
\begin{aligned} &\frac{x-3}{2}=-1 \\\\ &\frac{y-1}{-1}=-1 \\\\ &\frac{z-2}{1}=1\\ \end{aligned}
\begin{aligned} &x=1 \\ &y=2 \\ &z=1 \end{aligned}
Point$(1,2,1)$
Passing $(3,1,2),$ the point.

direction $\rightarrow 2,-1,1$
\begin{aligned} &\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1} \\\\ &\Rightarrow \vec{r}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\\\ &(2 \hat{\imath}+-\hat{\jmath}+\hat{k}) \end{aligned}
point $(3,1,2)$
point $q=(1,2,1)$
So the middle point $=2+\frac{3}{2}+\frac{3}{2}$

i.e, $2 i+\frac{3}{2} \hat{\jmath}+\frac{3}{2} k$

The Plane exercise 28.15 question 5

Answer: $\left(-\frac{1}{12}, \frac{25}{12},-\frac{1}{6}\right), \frac{13}{\sqrt{24}}$
Hint:
Let point

directional ratio are perpendicular to the line
Given:
Find the coordinate of the foot of the perpendicular from the point $(1,1,2)$ to the plane $2x-2y+4z+5=0$ . Also find the length of the perpendicular
Solution:
Direction ratio of plane $\vec{n}=2,-2,4$

equation of line $\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=\lambda(\text { say })$
\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+1, \\\\ &z=4 \lambda+2 \end{aligned}
Put in $2 x-2 y+4 z+5=0$
$2(2 \lambda+1)-2(-2 \lambda+1)+4(4 \lambda+2)+5=0$
\begin{aligned} &4 \lambda+2+4 \lambda-2+16 \lambda+8+5=0 \\\\ &24 \lambda+13=0 \\\\ &\lambda=\frac{-13}{24} \end{aligned}
Putting the value of λ iner
\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+1 \\\\ &z=4 \lambda+2 \end{aligned}
\begin{aligned} &x=2\left(-\frac{13}{24}\right)+1 \\\\ &y=-2\left(\frac{-13}{24}\right)+1 \\\\ &z=4\left(\frac{-13}{24}\right)+2 \end{aligned}
\begin{aligned} &x=\frac{-13+12}{12} \\\\ &y=\frac{13+12}{12} \\\\ &z=\frac{-13+12}{6} \end{aligned}
Distance $=\left(-\frac{1}{12}, \frac{25}{12},-\frac{1}{6}\right)$
Given point $(1,1,2)$
Distance $=\sqrt{\left(1+\frac{1}{12}\right)^{2}+\left(1-\frac{25}{12}\right)^{2}+\left(2+\frac{1}{6}\right)^{2}}$
\begin{aligned} &=\sqrt{\frac{169}{144}+\frac{169}{144}+\frac{169}{36}} \\\\ &=13 \sqrt{\frac{1}{144}+\frac{1}{144}+\frac{1}{36}} \end{aligned}
\begin{aligned} &=\frac{13}{12} \sqrt{1+1+4} \\\\ &=\frac{13 \sqrt{6}}{12} \end{aligned}

The Plane exercise 28.15 question 6

Answer:$1$
Hint:
Let equation of line passing through $P(1,-2,3)$
Distance$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$
Given:
Find the distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured along a line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$
Solution:
Let the equation of line passing through $P(1,-2,3)$ is
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-31}{c} \\\\ &\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda \text { (say) } \\\\ &Q=(2 x+1,3 \lambda-2,-6 \lambda+3) \end{aligned}
Q Lies in the plane $x-y+z=5$
\begin{aligned} &2 \lambda+1-3 \lambda+2-6 \lambda+3=5 \\\\ &-7 \lambda=-1 \\\\ &\lambda=\frac{1}{7} \end{aligned}
\begin{aligned} Q &=\left(\frac{2}{7}+1, \frac{3}{7} \cdot-2,-\frac{6}{7}+3\right) \\\\ &=\left(\frac{9}{7}, \frac{-11}{7}, \frac{15}{7}\right) \end{aligned}
\begin{aligned} &P=(1,-2,3) \\\\ &P \theta=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(-\frac{11}{7}+2\right)^{2}+\left(\frac{15}{7}-3\right)^{2}} \end{aligned}
$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{4.9}}$
\begin{aligned} &=\sqrt{\frac{4+9+36}{49}} \\\\ &=\sqrt{\frac{49}{49}} \\\\ &=1 \text { unit } A n s \end{aligned}

The Plane exercise 28.15 question 7

Answer: $(5.2,6) ; \sqrt{11}$
Hint:
Let $M$ be the foot of the perpendicular of the point $P(2,3,7)$ in the plane $3x-y-z=7$
Given:
Find the coordinate of foot of the perpendicular from the point$(2,3,7)$ to the plane
$3x-y-z=7$. Also find the length of the perpendicular.
Solution:
Let $M$ be the foot of the perpendicular of the point $P (2,3,7)$ in the plane $3x-y-z=7$ then $PM$ is normal to the plane so the direction ratio of $PM$ are proportional to $3,-1,-1$
Let the coordination of $M$ be $(3r+2,-r+3,-r+7)$

Since $M$ lies in the plane $3x-y-z=7,$
$9r+6+r-3+r-7=7$
$11r=11$
$r=1$
Substituting this in coordinates of $M$ ,
$M=(3 r+4,-r+3,-r+\lambda)=(5,2,6)$
Now length of the perpendicular from $P$ onto the plane
\begin{aligned} &=\left|\frac{3(2)-3-7-7}{\sqrt{9+1+1}}\right| \\\\ &=2 \frac{11}{\sqrt{11}} \end{aligned}
$\frac{\sqrt{11} \times \sqrt{11}}{\sqrt{11}}=\sqrt{11} \; \mathrm{Ans}$

The Plane exercise 28.15 question 8

Answer: $(-3,5,2)$
Hint:
Formula ,
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y+y_{1}}{b}=\frac{z-z_{1}}{c} \\\\ &=\frac{-2\left(a x_{1}+b y_{1}+c z_{1}+d\right)}{\left(a^{2}+b^{2}+c^{2}\right)} \\\\ &\hat{k} \mathrm{P}(1,3,4) \end{aligned}

$2x-y+z+3=0$
Given:
Find the image of the point with $(1,3,4)$ in the plane $2x-y+z+3=0$
Solution:
\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right) \\\\ &a x+b y+z+d=0 \\\\ &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-31}{c}=\frac{-2\left(a x_{1}+b y_{1}+c_{1}+d\right)}{\left(a^{2}+b^{2}+c^{2}\right)} \end{aligned}
\begin{aligned} &\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{-1}=\frac{-2(2-3+4+3)}{4+1+1}=-2 \\\\ &x=-4+1 \\\\ &y=2+3 \\\\ &z=-2+4 \end{aligned}
point $=(1,3,4)$
\begin{aligned} &x=-3 \\ &y=5, \\ &z=2 \end{aligned}
Mirror image $(-3,5,2)$

The Plane exercise 28.15 question 9

Answer:$13$
Hint:
Distance between the point$=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}$
The coordinate of the point corresponding to the position vector
Given:
Find the distance of the point with position vector $-\hat{\imath}-5 \hat{\jmath}-10 \hat{k}$ from the point of intersection of the line $\vec{r}=(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(3 i+4 \hat{\jmath}+12 \hat{k})$ with the plane$\vec{r} \cdot(\hat{\imath}-j+k)=5$
Solution:
The given equation of the line is
\begin{aligned} &\vec{r}=(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}+12 \hat{k}) \\\\ &\Rightarrow \vec{r}=(2+3 \lambda) \hat{\imath}+(-1+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k} \end{aligned}
The coordinate of any point in this are of the form
$(2+3 \lambda) i+(-1+4 \lambda) j+(2+2 \lambda) \hat{k}$
Since this point lies on the plane $\vec{r} \cdot(i-\hat{\jmath}+\hat{k})=5$
\begin{aligned} &{[(2+3 \lambda) \hat{\imath}+(-1+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k}] \hat{1}-\hat{\jmath}+\hat{k}=5} \\\\ &\Rightarrow 2+3 \lambda+1-4 \lambda+2+2 \lambda-5=0 \\\\ &\Rightarrow \lambda=0 \end{aligned}
So the coordinate of the point are
\begin{aligned} &(2+3 \lambda,-1+4 \lambda, 2+2 \lambda) \\\\ &=(2+0,-1+0,2+0) \\\\ &=(2,-1,2) \end{aligned}.............(1)
The coordinate of the point corresponding to the position vector $-i-5 \hat{\jmath}-10 \hat{k}$ are $(-1,5,-10)$ … (2)
Distance between (1) and (2)
\begin{aligned} &\therefore \sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}} \\\\ &=\sqrt{9+16+144} \\\\ &=13 \end{aligned}

The Plane exercise 28.15 question 10

Answer: $\frac{13}{12} \sqrt{6},\left(-\frac{1}{12}, \frac{25}{12},-\frac{2}{12}\right)$
Hint:
For P as distance put
\begin{aligned} &\vec{r}=i+\hat{\jmath}+2 \hat{k} \text { in } \\\\ &\vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{k})+5=0 P(1,1,2) \end{aligned}

Given:
Find the length and foot of the perpendicular from the point $(1,1,2)$ to the plane $\vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{k})+5=0$
Solution:
Distance $=\frac{(\hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{i})+5}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}$
\begin{aligned} &=\frac{1-2+8+5}{\sqrt{1+4+16}} \\\\ &=\frac{12}{\sqrt{21}} \end{aligned}
Direction of $\vec{n}=(1,-2,4)$

equation of line $\frac{x-1}{1}=\frac{y-1}{-2}=\frac{3-2}{4}=\lambda(\text { say })$
\begin{aligned} &x=\lambda+1 \\\\ &y=2 \lambda+1, \\\\ &z=4 \lambda+2 \end{aligned}
$P u t\; \vec{r}=(\lambda+1) i+(-2 \lambda+1) \hat{\jmath}+(4 \lambda+2) \hat{k} \text { in } \vec{r} \cdot(\hat{\imath}-\hat{y}+4 \hat{k})+5=0$
\begin{aligned} &{[(\lambda+1) i+(-2 \lambda+1) \hat{\jmath}+(4 \lambda+2) \hat{k}] \cdot[\hat{l}-\hat{y}+4 \hat{k}]+5=0} \\\\ &\lambda+1+4 \lambda-2+16 \lambda+8+5=0 \\\\ &21 \lambda+12=0 \end{aligned}
\begin{aligned} &\lambda=-\frac{12}{21} \\\\ &x=\frac{-4}{7}+1 \\\\ &=\frac{3}{7} \end{aligned}
\begin{aligned} &y=\frac{8}{7}+1 \\\\ &=\frac{15}{7} \\\\ &z=\frac{-16}{7}+2 \end{aligned}
\begin{aligned} &=-\frac{2}{7} \\\\ &\left(\frac{3}{7}, \frac{15}{7}, \frac{-2}{7}\right) \text { Ans. } \end{aligned}

The Plane exercise 28.15 question 11

Answer:$(7,0,3)(-1,4,-1), \sqrt{6}$
Hint:
$\hat{N} \quad(3,2,1)$

If $M$ is lying on the plane and the normal vector as well $\hat{N}$
$(a x+b y+c z+d), \hat{N}=\frac{(a)}{\sqrt{a^{2}+b^{2}+i^{2}}} \hat{\imath}+\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \hat{\jmath}+\frac{c}{\sqrt{a^{2}+b^{2}} t^{2}} \hat{k}$
Given:
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point $P(3,2,1)$ from the plane $2x-y+z+1=0$. Find also the image of the point in the plane
Solution:
\begin{aligned} &\hat{N}=\frac{2}{\sqrt{6}} \hat{\imath}-\frac{\hat{\jmath}}{\sqrt{6}}+\frac{\hat{k}}{\sqrt{6}}\\ &\hat{N} \end{aligned}

\begin{aligned} &\therefore \frac{x-3}{\frac{2}{\sqrt{6}}}=\frac{y-2}{\left(-\frac{1}{\sqrt{6}}\right)}=\frac{3-1}{\left(\frac{1}{\sqrt{6}}\right)}=\lambda \\\\ &x=\frac{3+2 \lambda}{\sqrt{6}}, y=\frac{2-\lambda}{\sqrt{6}}, z=\frac{1+\lambda}{\sqrt{6}} \end{aligned}
\begin{aligned} &=2\left(3+\frac{2 \lambda}{\sqrt{6}}\right)-\left(2-\frac{\lambda}{\sqrt{6}}\right)+\left(1+\frac{\lambda}{\sqrt{6}}\right)+1=0 \\\\ &=6+\frac{4 \lambda}{\sqrt{6}}-2+\frac{\lambda}{\sqrt{6}}+2+\frac{\lambda}{\sqrt{6}}=0 \end{aligned}
\begin{aligned} &6+\frac{6}{\sqrt{6}}=0 \\\\ &\Rightarrow \lambda=-\frac{6 \sqrt{6}}{6}=-\sqrt{6} \\\\ &M[3-2,2+1,1-1] \end{aligned}
\begin{aligned} &\therefore M(1,3,0) \\ &\rightarrow \quad Q=\sqrt{2^{2}+1+1}=\sqrt{6} \\ &P(3,2,1) \end{aligned}

\begin{aligned} &\Rightarrow \frac{(3-a)}{(2 / \sqrt{6})}=\frac{(2-b)}{(-1 / \sqrt{6})}=\frac{(1-c)}{1 / \sqrt{6}}=2 \sqrt{6} \\\\ &\Rightarrow(3-a)=4 \text { or }-4 \\\\ &\Rightarrow-10 r+7 \end{aligned}
\begin{aligned} &\Rightarrow(2-b)=-2 \text { or } 2 \\\\ &\Rightarrow b=-4 \text { or } 0 \\\\ &\Rightarrow(1-c)=2 \text { or }-2 \end{aligned}
\begin{aligned} &\Rightarrow c=-1 \\\\ &p !(-1,4,-1) \text { or } \\\\ &p=7,0,3 \end{aligned}

The Plane exercise 28.15 question 13

Answer:$\frac{12}{29}, \frac{-18}{29}, \frac{24}{29}$
Hint:
Let cosines be $L,M,N.$
$\vec{r} \cdot \hat{n}=d .$
Given:
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x-3y+4z-6=0$
Solution:
$2x-3y+4z-6=0$
\begin{aligned} &2 x-3 y+4 z=6 \\\\ &\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29} \\\\ &=\frac{2}{\sqrt{29}} x-\frac{3}{\sqrt{29}} y+\frac{4}{\sqrt{29}} 3 \\\\ &=\frac{6}{\sqrt{29}} \end{aligned}
Direction cosines $=\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{4}},+\frac{4}{\sqrt{29}}$

The Plane exercise 28.15 question 14

Answer:$\left(0, \frac{5}{2}, 0\right) ; \sqrt{6}$
Hint:
Direction ratio on $(2,-2,4)$
Equation of line $=\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

$2 x-2 y+4 z-6=0$
Given:
Find the length and foot of the perpendicular from the point $(1,3/2,2)$ to the plane
$2x-2y+4z-6=0$
Solution:
$2x-2y+4z+5=0$
\begin{aligned} &D R^{\prime} \Delta=(2,-2,4) \\\\ &A B D R^{\prime} S=(2,-2,4) \end{aligned}
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-3}{c} \\\\ &\frac{x-1}{2}=\frac{y-3 / 2}{-2}=\frac{2-2}{4}=1 \end{aligned}
\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+\frac{3}{2} \\\\ &z=4 \lambda+2 \end{aligned}
\begin{aligned} &\Rightarrow 2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0 \\\\ &\Rightarrow 4 \lambda+2+4 \lambda-3+16 \lambda+8+5=0 \end{aligned}
\begin{aligned} &24 \lambda=-12 \\\\ &\lambda=-\frac{12}{24} \\\\ &=-\frac{1}{2} \end{aligned}
Foot of the perpendicular,
\begin{aligned} &x=2 \lambda+1 \\\\ &=2\left(\frac{-1}{2}\right)+1=0 \\\\ &y=-2\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{5}{2} \\\\ &z=4\left(\frac{-1}{2}\right)+2=0 \end{aligned}
Foot of the perpendicular $\left(0, \frac{5}{2}, 0\right)$

$A B=\sqrt{1+1+4}=\sqrt{6}$

The Plane exercise 28.15 question 15

Answer:$\left(3, \frac{7}{2}, \frac{11}{2}\right),(4,4,7)$
Hint:
Distance $\rightarrow \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}$
$\vec{d}=\vec{r} \cdot \vec{n}$

Given:
Find the position vector of the foot of the perpendicular and the perpendicular distance from the point P with position vector $2 \tilde{i}+3 \hat{\jmath}+4 \hat{k}$ to the plane $\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+3 \hat{k})-26=0$ . Also find the image of $P$ in the plane
Solution:
\begin{aligned} &\frac{x-2}{2}=\frac{y-3}{1}=\frac{z-4}{3}=t \\\\ &x=2+2 t \\\\ &y=3+t \\\\ &z=4+3 t \end{aligned}
\begin{aligned} &\Rightarrow[(2+2 t) \hat{i}+(3+t) \hat{j}+(4+3 t) \hat{k}][2 \hat{i}+\hat{j}+3 \hat{k}]=26 \\\\ &\Rightarrow 19+14 t=26 \\\\ &t=\frac{1}{2} \end{aligned}
Putting the value of t in equation 1
i.e,
\begin{aligned} & x=3, \\\\ &y=\frac{7}{2}, \\\\ &z=\frac{11}{2} \end{aligned}
$\mathrm{PQ} \Rightarrow(3-2)^{2}+\left(\frac{7}{2}-3\right)^{2}+\left(\frac{11}{2}-2\right)^{2}$
\begin{aligned} &=\sqrt{1+\left(\frac{1}{2}\right)^{2}+\left(\frac{11}{2}\right)^{2}}, \\\\ &=\sqrt{1+\frac{1}{4}+\frac{9}{4}} \end{aligned}
\begin{aligned} &=\sqrt{\frac{14}{4}} \\\\ &=\frac{\sqrt{14}}{2} \end{aligned}
Image ,
$x=2+2(1)=4$
\begin{aligned} &y=3+1=4 \\\\ &z=4+3(1)=7 \end{aligned}
Image $(4,4,7)$ Ans
Foot $\left(3, \frac{7}{2}, \frac{11}{2}\right)$ Ans Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Chapter-wise Solutions

1. What is a line?

A line is a collection of points that extends infinitely in opposite directions. It only has one dimension, which is length. Collinear points are points that are on the same line.

2. What do you mean by a point?

A point is a location in a plane that has no size, i.e. no width, no length and no depth.

3. Can I study only using this material?

As this material covers all concepts, students can use it to prepare for their exams.

4. What is a plane?

It is entirely contained within a single plane. A plane figure c can be formed using line segments, curves, or a combination of the two, i.e. line segments and curves.

5. How do RD Sharma Class 12 Solutions help with CBSE Board Exams?

How do RD Sharma Class 12 Solutions aid in CBSE Board Exam preparation?

RD Sharma Solutions for Class 12 includes a large number of questions for regular practise. This has made it easier to finish all of the questions in exams on time. Throughout the academic year, they are kept busy solving problems by the short-answer and multiple-choice questions.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

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Application Date:05 September,2024 - 20 September,2024