Question

A line $\ell$ 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 ${\ell }_2$ at a distance of $\sqrt{17}$ from the point of intersection of $\ell \text{and}{\ell }_1$ is (are)

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

Answer 1

Answer: (B), (C)

$(–1,–1,0)$

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\lambda$ (Equation of line l)
Equation of a line $l_1,\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$
Equation of a line $l_2,\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$
Direction ratio of a line l is given by, $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|$
$=-2\hat{i}+3\hat{j}-2\hat{k}$
Equation of a line l is $\frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=\lambda$
Point of intersection of $l\text{and}{l}_1$,
$-2\lambda =3+t$ ...(1)
$3\lambda =2t-1$ ...(2)
Put the value of $t,3\lambda =2(-2\lambda -3)-1$
$3\lambda =-4\lambda -6-1$
$7\lambda =-7$
$\lambda =-1$
Point of intersection is $(2,–3,2)$
So, $\sqrt{(3+2s-2{)}^2+(3+2s+3{)}^2+(2+s-2{)}^2}=\sqrt{17}$
$\Rightarrow 4s^2+4s+1+36+24s+4s^2+s^2=17$
$\Rightarrow 9s^2+28s+20=0$
$\Rightarrow s=-2,-\frac{10}{9}$
i.e., intersection points are $(–1,–1,0)\text{and}(\frac{7}{9},\frac{7}{9},\frac{8}{9})$