Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \frac{\mathrm{x}-3}{7}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}+1}{-2}=\lambda \\ & \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\ & \text { dr's of QP } \Rightarrow 7 \lambda-7,-\lambda+5,-2 \lambda
\end{aligned}$
Now
$\begin{aligned}
& (7 \lambda-7) \cdot 7-(-\lambda+5)+(2 \lambda) \cdot 2=0 \\ & 54 \lambda-54=0 \Rightarrow \lambda=1 \\ & \therefore \mathrm{P}=(10,1,-3) \\ & \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{PR}}=-7 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \text { Area } \left.=\left|\frac{1}{2}\right| \begin{array}{ccc}
\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 0 & -4 & 2 \\ -7 & -3 & 4
\end{array} \right\rvert\,=3 \sqrt{30}
\end{aligned}$ ^

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line…

Solution