Question

A line $l$ passing through the origin is perpendicular to the lines
$l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty <t<\infty$
$l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty <s<\infty$

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)

$(\frac{7}{9},\frac{7}{9},\frac{8}{9})$

$l_1:(3\hat{i}-\hat{j}+4\hat{k})+t(\hat{i}+2\hat{j}+2\hat{k})$
$l_2:(3\hat{i}+3\hat{j}+2\hat{k})+s(2\hat{i}+2\hat{j}+\hat{k})$

Direction ratios of line $l$ is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|=-2\hat{i}+3\hat{j}-2\hat{k}$

So, equation of line $l$ is
$\frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=\lambda$
Any point on line $l$ is $(-2\lambda ,3\lambda ,-2\lambda )$

Intersection point of $l$ and $l_1:$
$-2\lambda =3+t\cdots \left(1\right)$
$3\lambda =-1+2t\cdots \left(2\right)$
$-2\lambda =4+2t$
Solving equations $\left(1\right)$ and $\left(2\right)$, we get
$\lambda =-1$
$\therefore$ Point of intersection is $(2,-3,2)$

So, $\sqrt{(3+2s-2{)}^2+(3+2s+3{)}^2+(2+s-2{)}^2}=\sqrt{17}$
$\Rightarrow 9s^2+28s+20=0$
$\Rightarrow s=-2,\frac{-10}{9}$

$\therefore$ Required points are $(-1,-1,0)$ and $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$