Let the acute angle bisector of the two planes x - 2y - 2z + 1 = 0 and 2x - 3y - 6z + 1 = 0 be the plane P. Then which of the following points lies on P?

Solution

Maths sir · 3 steps

Write both bisectors with the ± formula, pick the acute one by sign, then test each point in it.

1. (x-2y-2z+1)/3 = ± (2x-3y-6z+1)/7

2. Two bisectors. The normals have lengths √(1+4+4) = 3 and √(4+9+36) = 7. The + sign gives x - 5y + 4z + 4 = 0. The - sign gives 13x - 23y - 32z + 10 = 0.
3. Choose the acute one. Both constant terms are positive. The dot product of the normals is 2 + 6 + 12 = 20 > 0, so the + sign gives the obtuse bisector and the - sign gives the acute one. Check: the normal (13, -23, -32) makes cos θ = 123/3√1722 ≈ 0.99 with (1, -2, -2), so the bisector is almost parallel to the first plane, which only happens for the acute bisector.

   Point	13x - 23y - 32z + 10	On P?
   (3, 1, -1/2)	39 - 23 + 16 + 10 = 42	no
   (-2, 0, -1/2)	-26 - 0 + 16 + 10 = 0	yes
   (0, 2, -4)	0 - 46 + 128 + 10 = 92	no
   (4, 0, -2)	52 - 0 + 64 + 10 = 126	no

4. So. P is 13x - 23y - 32z + 10 = 0, and only (-2, 0, -1/2) lies on it. The answer is (b).

The idea to keep. First make both constant terms positive. If the normals' dot product is positive, the + bisector is obtuse and the - bisector is acute. If it is negative, the roles swap.

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