Let $P$ be the foot of the perpendicular from the point $M(1,2,2)$ on the line $L:\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z-2}{2}$. Let the line $\overrightarrow{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$, $\lambda\in\mathbf{R}$, intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:
Step-by-Step Solution
Key Concept: Find $P$ (foot of perp from $M$ to $L$) and $Q$ (intersection of the second line with $L$) separately, then compute $|PQ|^2$.
Foot $P$ on $L$: general point $(t+1,-t-1,2t+2)$. $\overrightarrow{MP}=(t,-t-3,2t)$.
$\overrightarrow{MP}\cdot(1,-1,2)=0$: $t+(t+3)+4t=0 \Rightarrow 6t+3=0 \Rightarrow t=-\tfrac{1}{2}$.
$P=\left(\tfrac{1}{2},-\tfrac{1}{2},1\right)$.
For $Q$: second line $(-1+\lambda,1-\lambda,-2+\lambda)$ must equal $(t+1,-t-1,2t+2)$.
$t+1=-1+\lambda$, $-t-1=1-\lambda$, $2t+2=-2+\lambda$.
From the first two: $t+\lambda=−2+2\lambda-2$... Adding first and second: $0=0$ (consistent). Using first: $\lambda-t=2$; using third: $\lambda=2t+4$. So $2t+4-t=2 \Rightarrow t=-2$, $\lambda=0$.
$Q=(-1,1,-2)$.
$PQ^2=\left(\tfrac{1}{2}+1\right)^2+\left(-\tfrac{1}{2}-1\right)^2+(1+2)^2=\tfrac{9}{4}+\tfrac{9}{4}+9=\tfrac{27}{2}$.
$2(PQ)^2=27$.
Correct Answer: 4