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