The square of the distance of the point ((15)/(7),(32)/(7),7) from the line (x+1)/(3)=(y+3…

Question

The square of the distance of the point $(\frac{15}{7},\frac{32}{7},7)$ 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:

Accepted Answer

$$\begin{aligned}
& L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \
& P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda
\end{aligned}$$

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

Solutions · 1