JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 17)
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 :
$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is :
102
138
132
108
Explanation
_1st_February_Evening_Shift_en_17_1.png)
$\begin{aligned} & \overrightarrow{\mathrm{PN}}. \overrightarrow{\mathrm{b}}=0\\\\ & 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \\\\ & 14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}\end{aligned}$
$\begin{aligned} & \mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right) \\\\ & \therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}\end{aligned}$
$\begin{aligned} & \frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \\\\ & \frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}\end{aligned}$
Now, $14(\alpha+\beta+\gamma)=14\left(\frac{62+4-12}{7}\right)=108$
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