Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{-2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:
Step-by-Step Solution
Key Concept: Find the direction of $L$ as the cross product of the two given direction vectors, write parametric equations of $L$, set $x=0$ for the $yz$-plane intersection, then compute $|PQ|$.
Direction of $L$: $(2,1,-2)\times(1,3,4)=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-2\\1&3&4\end{vmatrix}=(4+6)\hat{i}-(8+2)\hat{j}+(6-1)\hat{k}=10\hat{i}-10\hat{j}+5\hat{k}=5(2,-2,1)$.
Parametric: $(2+2\lambda,-1-2\lambda,3+\lambda)$.
$yz$-plane: $x=0 \Rightarrow 2+2\lambda=0 \Rightarrow \lambda=-1$.
$Q=(0,1,2)$.
$PQ=\sqrt{(2-0)^2+(-1-1)^2+(3-2)^2}=\sqrt{4+4+1}=3$.
Correct Answer: 4