If the mirror image of the point P(3,4,9) in the line (x-1)/(3)=(y+1)/(2)=(z-2)/(1) is (, …

Question

If the mirror image of the point $P(3,4,9)$ in the line 
$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is :

Accepted Answer

$\begin{aligned} & \overrightarrow{\mathrm{PN}}. \overrightarrow{\mathrm{b}}=0\\ & 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \\ & 14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}\end{aligned}$

$\begin{aligned} & \mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}ight) \\ & 	herefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}\end{aligned}$

$\begin{aligned} & \frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \\ & \frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}\end{aligned}$

Now, $14(\alpha+\beta+\gamma)=14\left(\frac{62+4-12}{7}ight)=108$

Solutions · 1