Let the image of the point P(2,-1,3) in the plane x+2 y-z=0 be Q.

Question

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 :

Accepted Answer

Equation 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 ' $m$ ' $(\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0$

$$ \Rightarrow $$ $$
\lambda=\frac{1}{2}
$$

Point $\mathrm{m}\left(\frac{1}{2}+2,2 \times \frac{1}{2}-1, \frac{-1}{2}+3\right)$

$$
=\left(\frac{5}{2}, 0, \frac{5}{2}\right)
$$

For Image $\mathrm{Q}(\alpha, \beta, \gamma)$

$$
\begin{aligned}
& \frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0, \\
& \frac{\gamma+3}{2}=\frac{5}{2}
\end{aligned}
$$

$$
\begin{aligned}
& Q:(3,1,2) \\
& d=\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right| \\
& \Rightarrow d=\frac{42}{\sqrt{14}}=3 \sqrt{14}
\end{aligned}
$$

Solutions · 1