The distance of the point (1, –2, 3) from
the plane x – y + z = 5 measured parallel to
the line x/2 = y/3 = z/(- 6) is :

Solution

Maths sir · 4 steps

Do not drop a perpendicular here. Walk from the point along the given direction until you hit the plane.

1. Line through the point. The direction to move along is (2, 3, -6). A general point on the line through (1, -2, 3) in that direction is (1 + 2t, -2 + 3t, 3 - 6t).
2. Meet the plane. Put this point into x - y + z = 5: (1 + 2t) - (-2 + 3t) + (3 - 6t) = 5. This gives 6 - 7t = 5, so t = 1/7. The meeting point is (9/7, -11/7, 15/7).
3. Length of the move. The distance moved is |t| times the length of the direction vector, and √(2² + 3² + (-6)²) = √49 = 7.

   d = |t| × √(2² + 3² + 6²) = 1/7 × 7 = 1

4. So. The distance measured parallel to the line is 1, which is option (b).

The idea to keep. For distance measured along a given line, write the point as P + td⃗, find t at the plane, and take |t| |d⃗|. The parameter t is a distance only when d⃗ is a unit vector.

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