Let d be the distance between the foot of perpendiculars of the points P(1, 2, -1) and Q(2, -1, 3) on the plane -x + y + z = 1. Then d² is equal to ___________.

Solution

Maths sir · 4 steps

Check PQ against the normal first: here it is perpendicular, so the feet are as far apart as P and Q.

1. Projection idea. The segment joining the two feet is the projection of PQ⃗ onto the plane. So remove the part of PQ⃗ along the normal n⃗ = (-1, 1, 1), and what is left has length d.

   d² = |PQ⃗|² - (PQ⃗·n⃗)²/|n⃗|²

2. Put in numbers. PQ⃗ = (1, -3, 4), so |PQ⃗|² = 1 + 9 + 16 = 26. Also PQ⃗·n⃗ = -1 - 3 + 4 = 0, so PQ is parallel to the plane and nothing is removed.
3. Cross-check feet. Both points give -x + y + z = 0, so each foot is 1/3n⃗ away: (2/3, 7/3, -2/3) and (5/3, -2/3, 10/3). Their difference is (1, -3, 4), the same as PQ⃗.
4. So. d² = 26 - 0, so d² is 26.

The idea to keep. The distance between two feet of perpendiculars is the projection of PQ onto the plane: d² = |PQ|² - (PQ· n)²/|n|². If PQ· n = 0, the answer is just |PQ|².

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