3D Geometry 3 Question 7
####7. If $Q(0,-1,-3)$ is the image of the point $P$ in the plane $3 x-y+4 z=2$ and $R$ is the point $(3,-1,-2)$, then the area (in sq units) of $\triangle P Q R$ is
(a) $\frac{\sqrt{91}}{2}$
(b) $2 \sqrt{13}$
(c) $\frac{\sqrt{91}}{4}$
(d) $\frac{\sqrt{65}}{2}$
(2019 Main, 10 April I)
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Answwer:
(a)
Solution:
- Given, equation of plane $3 x-y+4 z=2$…(i) and the point $Q(0,-1,-3)$ is the image of point $P$ in the plane (i), so point $P$ is also image of point $Q$ w.r.t. plane (i).
Let the coordinates of point $P$ is $\left(x_{1}, y_{1}, z_{1}\right)$, then
$ \begin{aligned} \frac{x_{1}-0}{3} & =\frac{y_{1}+1}{-1}=\frac{z_{1}+3}{4} \\ & =-2 \frac{[3(0)-1(-1)+4(-3)-2]}{3^{2}+(-1)^{2}+4^{2}} \end{aligned} $
$\left[\because\right.$ image of the point $\left(x_{1}, y_{1}, z_{1}\right)$ in the plane $a x+b y+c z+d=0$ is $(x, y, z)$, where
$ \begin{aligned} & \left.\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)}{a^{2}+b^{2}+c^{2}}\right] \\ & \Rightarrow \quad \frac{x_{1}-0}{3}=\frac{y_{1}+1}{-1}=\frac{z_{1}+3}{4} \\ & \quad=2 \frac{(1-12-2)}{26}=\frac{26}{26}=1 \\ & \Rightarrow \quad P\left(x_{1}, y_{1}, z_{1}\right)=(3,-2,1) \end{aligned} $
Now, area of $\triangle P Q R$, where point $R(3,-1,-2)$
$ \begin{aligned} =\frac{1}{2}|\overrightarrow{P \vec{Q}} \times P \vec{R}| & =\frac{1}{2}|(-3 \hat{i}+\hat{j}-4 \hat{k}) \times(0 \hat{i}+\hat{j}-3 \hat{k})| \\ & =\frac{1}{2}\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ -3 & 1 & -4 \\ 0 & 1 & -3 \end{array}\right|=\frac{1}{2}|\hat{i}-9 \hat{j}-3 \hat{k}| \\ & =\frac{1}{2} \sqrt{1+81+9}=\frac{\sqrt{91}}{2} \text { squnits } \end{aligned} $
