JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Three Dimensional Geometry

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

, is:

(A) $116$
(B) $11/ 6$ ✓ Correct
(C)11
(D) $611$

Answer: (B)

Solution

The plane holds both lines, so it passes through the point

(- 2, 2, - 5)

on the first line and contains the two direction vectors

(3, 5, 7)

and

(1, 4, 7)

. Its normal is their cross product,

(3, 5, 7) \times (1, 4, 7) = (7, - 14, 7) = 7 (1, - 2, 1)

. Using the point, the plane is

1 (x + 2) - 2 (y - 2) + 1 (z + 5) = 0

, i.e.

x - 2 y + z + 11 = 0

. The distance from the origin is

\frac{∣11∣}{1 + 4 + 1} = \frac{11}{6}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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