Three Dimensional Geometry Question 4
Question 4 - 2024 (01 Feb Shift 2)
If the mirror image of the point $\mathrm{P}(3,4,9)$ 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 :
(1) 102
(2) 138
(3) 108
(4) 132
Show Answer
Answer (3)
Solution
$\overrightarrow{\mathrm{PN}} \overrightarrow{\mathrm{b}}=0$ ?
$3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0$
$14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}$
$\mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)$
$\therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}$
$\frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7}$
$\frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}$
Ans. $14(\alpha+\beta+r)=108$
