If the four points, whose position vectors are 3î - 4ĵ + 2k̂,î + 2ĵ - k̂, - 2î - ĵ + 3k̂ and 5î - 2α ĵ + 4k̂ are coplanar, then α is equal to :

Solution

Maths sir · 4 steps

Four points are coplanar exactly when the three edge vectors from one of them have a zero triple product.

1. Vectors from A. Call the points A(3,-4,2), B(1,2,-1), C(-2,-1,3) and D(5,-2α,4). Then AB⃗ = (-2,6,-3), AC⃗ = (-5,3,1) and AD⃗ = (2, 4-2α, 2).
2. Coplanarity condition. All four points lie in one plane only if these three vectors lie in one plane. So their scalar triple product, the determinant below, must be zero. Write t = 4-2α to keep the working short.

   |-2, 6, -3; -5, 3, 1; 2, t, 2| = -2(6-t) - 6(-10-2) - 3(-5t-6) = 78 + 17t

3. Solve for t. Setting 78 + 17t = 0 gives t = -78/17. Then 4 - 2α = -78/17, so 2α = 68/17 + 78/17 = 146/17.
4. So. α = 73/17, which is option (a).

The idea to keep. For four coplanar points, take differences from one point to get three vectors and set their determinant to zero. Use differences, not the position vectors themselves.

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