Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$ and $\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$…

Question

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$ and $\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :

Aicharya · Accepted Answer

Vector parallel to $L$
$\begin{aligned}
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4
\end{array}\right|=10 \hat{i}-10 \hat{j}+5 \hat{k} \\ & =5(2 \hat{i}-2 \hat{j}+\hat{k})
\end{aligned}$
Equation of ' $L$ '
$\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-3}{1}=\lambda(\text { say })$
Let $Q(2 \lambda+2,-2 \lambda-1, \lambda+3)$
$\begin{aligned}
& \Rightarrow \quad 2 \lambda+2=0 \Rightarrow \lambda=-1 \\ & \Rightarrow Q(0,1,2) \\ & d(P, Q)=3
\end{aligned}$ .

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines…

Solution