Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is…

Question

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :

Aicharya · Accepted Answer

General point on line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$
is $(\lambda+1,-\lambda-1,2 \lambda+2)$

DR's of PM are $(\lambda,-\lambda-3,2 \lambda)$
$P M \perp L$
$\begin{aligned}
& \Rightarrow \lambda+(-1)(-\lambda-3)+2(2 \lambda)=0 \\ & \Rightarrow 6 \lambda+3=0
\end{aligned}$

$P\left(\frac{1}{2}, \frac{-1}{2}, 1\right)$
Let another line $L^{\prime}: \frac{x+1}{1}=\frac{y-1}{-1}=\frac{z+2}{1}$
General point on line $L^{\prime}$ is $(\mu-1,-\mu+1, \mu-2)$
Point of intersection of line $L$ and $L^{\prime}$ is
$\begin{array}{l|l}
\lambda+1=\mu-1 & 2 \lambda+2=\mu-2 \\ \Rightarrow \mu-\lambda=2 \ldots(1) & \Rightarrow 2 \lambda=\mu-4
\end{array}$

$\begin{aligned} & Q(-1,1,-2) \\ & 2(P Q)^2=2\left(\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2\right) \\ & =2\left(\frac{9}{4}+\frac{9}{4}+9\right) \\ & =27\end{aligned}$ *

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}:…

Solution