Three Dimensional Geometry Question 32
Question 32 - 01 February - Shift 1
Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$. Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is
(1) $\frac{22 \sqrt{2}}{7}$
(2) $\frac{24 \sqrt{2}}{7}$
(3) $2 \sqrt{14}$
(4) $3 \sqrt{14}$
Show Answer
Answer: (4)
Solution:
Formula: A Plane and A Point (7.1) and (7.4)
eq. of line PM $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda$
any point on line $=(\lambda+2,2 \lambda-1,-\lambda+3)$
for point ’ $m$ ’ $(\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0$
$ \lambda=\frac{1}{2} $
Point $m(\frac{1}{2}+2,2 \times \frac{1}{2}-1, \frac{-1}{2}+3)$
$=(\frac{5}{2}, 0, \frac{5}{2})$
For Image Q $(\alpha, \beta, \gamma)$
$\frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0$,
$\frac{\gamma+3}{2}=\frac{5}{2}$
$Q:(3,1,2)$
$d=|\frac{3(3)+2(1)+2+29}{\sqrt{3^{2}+2^{2}+1^{2}}}|$
$d=\frac{42}{\sqrt{14}}=3 \sqrt{14}$
