Let p be the plane passing through the intersection of the planes r⃗ . ( î + 3ĵ - k̂ ) = 5 and r⃗ . ( 2î - ĵ + k̂ ) = 3, and the point (2, 1, -2). Let the position vectors of the points X and Y be î - 2ĵ + 4k̂ and 5î - ĵ + 2k̂ respectively. Then the points :

Solution

Maths sir · 3 steps

Pin down the plane first, then put each point into its equation: the sign tells you the side.

1. Family of planes. Any plane through the intersection of the two given planes has the form (x + 3y - z - 5) + λ(2x - y + z - 3) = 0. At (2, 1, -2) the first bracket is 2 and the second is -2, so 2 - 2λ = 0 and λ = 1.

   P: 3x + 2y - 8 = 0

2. Side test. Two points are on the same side of P when 3x + 2y - 8 has the same sign at both. Opposite signs mean opposite sides. Find each point's coordinates first, then substitute.

   Point	Coordinates	3x + 2y - 8	Sign
   X	(1, -2, 4)	3 - 4 - 8 = -9	negative
   Y	(5, -1, 2)	15 - 2 - 8 = 5	positive
   X + Y	(6, -3, 6)	18 - 6 - 8 = 4	positive
   Y − X	(4, 1, -2)	12 + 2 - 8 = 6	positive
   X − Y	(-4, -1, 2)	-12 - 2 - 8 = -22	negative

3. So. X gives -9 and Y gives 5, so X and Y are on opposite sides of P. Each of the other three options fails a sign check in the table.

The idea to keep. To test which side of a plane a point lies on, substitute it into ax + by + cz + d. Only the sign matters. For combinations like X + Y, find the new point first and then substitute.

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