The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the points A, B, C is (2, a, 4), a ∈ N. If the volume of the tetrahedron OABC is 144 unit³, then which of the following points is NOT on P ?

Solution

Maths sir · 4 steps

The foot of the perpendicular from O gives you the normal and a point on the plane together.

1. Write the plane. The line from O to the foot (2, a, 4) is perpendicular to P, so (2, a, 4) is a normal to P. The foot also lies on P, so P is 2x + ay + 4z = 4 + a² + 16 = 20 + a².
2. Find the intercepts. Let k = 20 + a². Then P meets the axes at OA = k/2, OB = k/a and OC = k/4. The tetrahedron OABC has volume 1/6 × OA × OB × OC.

   1/6 × k/2 × k/a × k/4 = (20 + a²)³/48a = 144 ⇒ (20 + a²)³ = 6912a

3. Solve for a. Try natural numbers. For a = 1: 21³ = 9261 ≠ 6912. For a = 2: 24³ = 13824 = 6912 × 2. So a = 2, and P is 2x + 2y + 4z = 24, that is, x + y + 2z = 12.

   Point	x + y + 2z	On P?
   (3, 0, 4)	3 + 0 + 8 = 11	No
   (0, 6, 3)	0 + 6 + 6 = 12	Yes
   (0, 4, 4)	0 + 4 + 8 = 12	Yes
   (2, 2, 4)	2 + 2 + 8 = 12	Yes

4. So. Only (3, 0, 4) fails to satisfy x + y + 2z = 12, so the answer is (a).

The idea to keep. When you know the foot of the perpendicular from the origin to a plane, that point is the normal vector and a point on the plane. Write the plane at once as r⃗ · n⃗ = |n⃗|².

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