Let $L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$, $\alpha\in\mathbf R$, 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 _______

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] On $L_1$: $z=-1$, $y=1-s$, $x=1+3s$; $L_2$ has $y=0$, so $s=1$ and $B=(4,0,-1)$. On $L_2$ at $B$: $t=1$ and $-4+\alpha=-1$, so $\alpha=3$. The foot of the perpendicular from $A$ on $L_2$ (direction $(2,0,3)$) is at $t=\frac7{13}$: $P=\left(\frac{40}{13},0,-\frac{31}{13}\right)$. $PB^2=\frac{144+324}{169}=\frac{36}{13}$, so $26\cdot3\cdot\frac{36}{13}=216$. Answer: 216.

Q. Let $L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$, $\alpha\in\mathbf R$, 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 _______

Explanation

Reviewed by Dr. Abubacker

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