Question

A line $l$ passing through the origin is perpendicular to the lines
$l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty <t<\infty$,
$l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty <t<\infty$,
Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$is (are)

• A. $(\frac{7}{3},\frac{7}{3},\frac{5}{3})$
• B. $(-1,-1,0)$
• C. $(1,1,1)$
• D. $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$

Answer 1

Answer: (B), (D)

$(-1,-1,0)$
$(\frac{7}{9},\frac{7}{9},\frac{8}{9})$

The common perpendicular is along $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|=$ $-2\hat{i}+3\hat{j}-2\hat{k}$
Let $M\equiv (2\lambda ,-3\lambda ,2\lambda )$
$So,\frac{2\lambda -3}{1}=\frac{-3\lambda +1}{2}=\frac{2\lambda -4}{2}=>\lambda =1$
So, $M\equiv (2,-3,2)$
Let the required point be $P$
Given that $PM=\sqrt{17}$
$=>(3+2s-2{)}^2+(3+2s+3{)}^2+(2+s-2{)}^2=17$
$=>9s^2+28s+20=0$
$=>s=-2,-\frac{10}{9}$
So, $P\equiv (-1,-1,0)$ or $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$
.