The distance of the point (7, -3, -4) from the plane passing through the points (2, -3, 1), (-1, 1, -2) and (3, -4, 2) is :

Solution

Maths sir · 4 steps

Two edge vectors in the plane give the normal by cross product, then use the point-to-plane formula.

1. Two vectors in plane. Call the points A(2,-3,1), B(-1,1,-2) and C(3,-4,2). Then AB⃗ = (-3, 4, -3) and AC⃗ = (1, -1, 1) both lie in the plane.

   n⃗ = AB⃗ × AC⃗ = |î, ĵ, k̂; -3, 4, -3; 1, -1, 1| = (4-3) î - (-3+3) ĵ + (3-4) k̂ = î - k̂

2. Equation of plane. With normal (1, 0, -1) through A: (x-2) - (z-1) = 0, that is x - z - 1 = 0. Check with B: -1 + 2 - 1 = 0, and with C: 3 - 2 - 1 = 0.
3. Distance formula. For (7,-3,-4): d = (|7 - (-4) - 1|)/√(1² + 0² + (-1)²) = 10/√2.
4. So. d = 10/√2 = 5√2, which is option (d).

The idea to keep. Three points fix a plane: cross two edge vectors to get the normal, then distance is (|ax₁+by₁+cz₁+d|)/√(a²+b²+c²). Always check the plane with a third point.

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