A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (α, –3, 5), then α is equal to:

Solution

Maths sir · 4 steps

Both direction vectors lie in the plane, so their cross product is the normal. Find it, then substitute.

1. Find the normal. A line lying in the plane is perpendicular to the plane's normal. The normal is perpendicular to both given directions, so it is their cross product.

   n⃗ = |î, ĵ, k̂; 1, -2, 2; 2, 3, -1| = (2-6) î - (-1-4) ĵ + (3+4) k̂ = -4î + 5ĵ + 7k̂

2. Equation of plane. Through (3, 1, 1) with normal (-4, 5, 7): -4(x-3) + 5(y-1) + 7(z-1) = 0. This simplifies to -4x + 5y + 7z = 0. Check: -12 + 5 + 7 = 0.
3. Substitute the point. Put (α, -3, 5) into the plane: -4α - 15 + 35 = 0, so 4α = 20.
4. So. α = 5, which is option (c).

The idea to keep. Two directions lying in a plane? Their cross product is the normal. One point plus the normal fixes the plane completely.

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