Let P be the foot of the perpendicular from the point (1,2,2) on the line L:(x-1)/(1)=(y+1…

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu \
& \mathrm{P}(\mu+1,-\mu-1,2 \mu+2) \
& \overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{~d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0 \
& \Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2} \
& \therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right) \
& \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right)
\end{aligned}$$

$$\[\begin{array}{c|c|c} \mu+1=-1+\lambda & -\mu-1=1-\lambda & 2 \mu+2=-2+\lambda \ \mu=\lambda-2 & \mu=\lambda-2 & \downarrow \end{array}$$\]

$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\begin{aligned}
& 2(\lambda-2)+2=-2+\lambda \
& 2 \lambda-4+2=-2+\lambda
\end{aligned}$$

$$\begin{aligned}
& \therefore \quad \mu=-2, \lambda=0 \
& \therefore \quad \mathrm{Q} \equiv(-1,1-2) \
& \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \text { and } \mathrm{Q}(-1,1,-2) \
& \mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2} \
& =\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}} \
& \therefore \quad 2(\mathrm{PQ})^2=2\left(\frac{54}{4}\right)=27
\end{aligned}$$

Solutions · 1