Let $L_1:\dfrac{x-1}{3}=\dfrac{y-1}{-1}=\dfrac{z+1}{0}$ and $L_2:\dfrac{x-2}{2}=\dfrac{y}{0}=\dfrac{z+4}{\alpha}$, $\alpha\in\mathbf{R}$, be two lines which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26\alpha(PB)^2$ is ________.
Step-by-Step Solution
Key Concept: Find the intersection $B$ by equating parametric forms of $L_1$ and $L_2$ to get $\lambda,\mu,\alpha$; then find $P$ as the foot of perpendicular from $A$ to $L_2$ using the dot product condition.
Parametrically: $L_1=(3\lambda+1,1-\lambda,-1)$, $L_2=(2\mu+2,0,\alpha\mu-4)$.
Equating: $3\lambda+1=2\mu+2$, $-\lambda+1=0 \Rightarrow \lambda=1$, $-1=\alpha\mu-4$.
From $\lambda=1$: $4=2\mu+2 \Rightarrow \mu=1$; then $-1=\alpha-4 \Rightarrow \alpha=3$. $B=(4,0,-1)$.
Foot $P$ on $L_2=(2\delta+2,0,3\delta-4)$: $\overrightarrow{AP}\cdot(2,0,3)=0$.
$(2\delta+1)\cdot2+(3\delta-3)\cdot3=0 \Rightarrow 13\delta=7 \Rightarrow \delta=\tfrac{7}{13}$.
$P=\left(\tfrac{40}{13},0,-\tfrac{31}{13}\right)$.
$PB^2=\left(4-\tfrac{40}{13}\right)^2+\left(-1+\tfrac{31}{13}\right)^2=\left(\tfrac{12}{13}\right)^2+\left(\tfrac{18}{13}\right)^2=\dfrac{144+324}{169}=\dfrac{468}{169}$.
$26\alpha(PB)^2=26\times3\times\dfrac{468}{169}=78\times\dfrac{468}{169}=216$.
Correct Answer: 216