Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, 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…

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Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, 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 $26 \alpha(\mathrm{~PB})^2$ is _________

Aicharya · Accepted Answer

Point $B$
$\begin{aligned}
& (3 \lambda+1,-\lambda+1,-1) \equiv(2 \mu+2,0, \alpha \mu-4) \\ & 3 \lambda+1=2 \mu+2 \\ & -\lambda+1=0 \\ & -1=\alpha \mu-4 \\ & \lambda=1, \mu=1, \alpha=3 \\ & B(4,0,-1)
\end{aligned}$
$\text { Let Point ' } P \text { ' is }(2 \delta+2,0,3 \delta-4)$
$\text { Dr's of AP } < 2 \delta+1,-1,3 \delta-3>$
$A P \perp L_2 \Rightarrow \delta=\frac{7}{13}$
$P\left(\frac{40}{13}, 0, \frac{-31}{13}\right)$
$\therefore 26 \alpha(P B)^2=26 \times 3 \times\left(\frac{144}{169}+\frac{324}{169}\right)$
$=216$

Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2:…

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