JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Three Dimensional Geometry

Let

P_{1}

and

P_{2}

be two planes given by

P_{1} : 10 x + 15 y + 12 z - 60 = 0,

P_{2} : - 2 x + 5 y + 4 z - 20 = 0.

Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on

P_{1}

and

P_{2}

Answer: (A), (B), (D)

Solution

The line of intersection of

P_{1}

and

P_{2}

can be found. The direction vector is

n_{1} \times n_{2}

where

n_{1} = (10, 15, 12)

and

n_{2} = (- 2, 5, 4)

.

A line can be an edge of a tetrahedron with two faces on

P_{1}

and

P_{2}

if:

The line of intersection passes through a point on both planes and has direction proportional to

(0, - 4, 5)

(after calculation), which can be written as

\frac{x}{0} = \frac{y - 4}{- 4} = \frac{z}{5}

.

Checking each option:

(A) is parallel to the z-axis and skew to the intersection line ✓
(B) is skew to the intersection line ✓
(C) lies on $P_{2}$ (verify: the line satisfies $P_{2}$ ) but passes through the intersection line, so it cannot form a tetrahedron edge ✗
(D) is skew to the intersection line ✓

Options (A), (B), (D) are correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options