JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Three Dimensional Geometry

If the straight lines

\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}

and

\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}

are coplanar, then the plane(s) containing these two lines is(are)

Answer: (B), (C)

Solution

For the lines to be coplanar, the determinant condition gives:

252 k 20 2 k 0 = 0

Expanding:

2 (0 - 0) - 0 + 0 = 2 (k \cdot 0 - 2 k) = - 4 k

... Actually, expanding along the third row:

2 k 2 2 k = 2 (k^{2} - 4) = 0

k^{2} = 4

, giving

k = \pm 2

.

For

k = 2

: The lines pass through

(1, - 1, 0)

and

(- 1, - 1, 0)

with directions

(2, 2, 2)

and

(5, 2, 2)

. The plane containing them satisfies

y + 1 = z

, or

y - z = - 1

✓ (C)

For

k = - 2

: The plane is

y + z = - 1

✓ (B)

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · One or more correct options