Let (x - 2)/3 = (y + 1)/(- 2) = (z + 3)/(- 1) lie on the plane px - qy + z = 5, for some p, q ∈ R. The shortest distance of the plane from the origin is :

Solution

Maths sir · 4 steps

A line sits in a plane when one point lies on it and its direction is perpendicular to the normal.

1. Read the line. The line passes through (2, -1, -3) and has direction (3, -2, -1). The plane px - qy + z = 5 has normal (p, -q, 1). Keep the minus sign on q.
2. Point on plane. Put (2, -1, -3) into the plane: 2p + q - 3 = 5, so 2p + q = 8.
3. Direction meets normal. The direction must be perpendicular to the normal, so their dot product is zero: 3p + 2q - 1 = 0, so 3p + 2q = 1. Substitute q = 8 - 2p to get -p + 16 = 1. So p = 15 and q = -22.

   d = (|0 + 0 + 0 - 5|)/√(15² + 22² + 1²) = 5/√710

4. So. The plane is 15x + 22y + z = 5, and 5/√710 = √(25/710) = √(5/142). The answer is (b).

The idea to keep. For a line to lie in a plane you need two conditions: one point of the line satisfies the plane, and the line's direction dotted with the plane's normal is zero. Two conditions fix two unknowns.

Answer from the NTA final answer key. Solution by PrepWiser.