The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

Question

Aicharya · Accepted Answer

$\begin{aligned} & L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\ & P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda\end{aligned}$
$\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)$
Since Q lies on line L
$\begin{aligned}
& \text { So, } \frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7} \\ & \Rightarrow 7 \lambda+22=21 \lambda+36 \\ & \Rightarrow \lambda=-1 \\ & \therefore \text { Point } Q\left(\frac{8}{7}, \frac{4}{7}, 0\right) \\ & \mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^2+\left(\frac{32}{7}-\frac{4}{7}\right)^2+(7-0)} \\ & \mathrm{PQ}=\sqrt{66} \\ & \Rightarrow(\mathrm{PQ})^2=66
\end{aligned}$ .

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line…

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