A line $l$ passing through the origin is perpendicular to the lines \[ \begin{aligned} & l_{1}:(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k},-\infty<t<\infty \\ & l_{2}:(3+2 s) \hat{i}+(3+2 s) \hat{j}+(2+s) \hat{k},-\infty<s<\infty \end{aligned} \] 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)

(B) (-1, -1, 0); (D) $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

(A) $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Three Dimensional Geometry

A line

l

passing through the origin is perpendicular to the lines

l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty

Then, the coordinate(s) of the point(s) on

l_{2}

at a distance of

17

from the point of intersection of

l

and

l_{1}

is (are)

Answer: (B), (D)

The direction vectors of

l_{1}

and

l_{2}

are

(1, 2, 2)

and

(2, 2, 1)

respectively. A line perpendicular to both has direction vector

(1, 2, 2) \times (2, 2, 1) = (2, - 3, 2)

. So

l

\frac{x}{2} = \frac{y}{- 3} = \frac{z}{2}

. The intersection of

l

and

l_{1}

is found by solving, giving point

A = (2, - 3, 2)

. A general point on

l_{2}

B = (3 + 2 s, 3 + 2 s, 2 + s)

. The distance

A B = 17

gives

(2 s + 1)^{2} + (2 s + 6)^{2} + s^{2} = 17

, which simplifies to

9 s^{2} + 28 s + 20 = 0

. Solving:

s = - 2

s = - \frac{10}{9}

. For

s = - 2

B = (- 1, - 1, 0)

. For

s = - \frac{10}{9}

B = (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · One or more correct options