Let a line L pass through the point P(2,3,1) and be parallel to the line x+3 y-2 z-2=0=x-y…

Question

Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to :

Answer: 158

Accepted Answer

$L:\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-1}{-1}=\lambda $

Any point on L can be taken as

$B(\lambda +2,-\lambda +3,-\lambda +1)$

Let $A(5,3,8)$

So, $AB\, .\, (\hat{i}-\hat{j}-\hat{k})=0$

$[(\lambda -3)\hat{i}-\lambda \hat{j}-(\lambda +7)\hat{k}]\, .\, [\hat{i}-\hat{j}-\hat{k}]=0$

$\lambda -3+\lambda +\lambda +7=0$

$	herefore $ $\lambda =\frac{-4}{3}$

$\overset{	o \leftarrow }{AB}=\frac{13}{3}\hat{i}+\frac{4}{3}\hat{i}-\frac{17}{3}\hat{k}$

$|\overset{	o \leftarrow }{AB}|=\sqrt{\frac{169}{9}+\frac{16}{9}+\frac{289}{9}}$

$=\frac{\sqrt{474}}{3}=\alpha $

$3{\alpha }^{2}=\frac{474}{9}	imes 3=158$

Solutions · 1