The sum of all values of α, for which the points whose position vectors are î-2 ĵ+3 k̂, 2 î-3 ĵ+4 k̂,(α+1) î+2 k̂ and 9 î+(α-8) ĵ+6 k̂ are coplanar, is equal to :

Solution

Maths sir · 4 steps

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

1. Vectors from A. Call the points A(1,-2,3), B(2,-3,4), C(α+1,0,2) and D(9,α-8,6). Then AB⃗ = (1,-1,1), AC⃗ = (α,2,-1) and AD⃗ = (8,α-6,3).
2. Coplanarity condition. The four points lie in one plane exactly when these three vectors lie in one plane. That happens when their scalar triple product, the determinant below, is zero.

   |1, -1, 1; α, 2, -1; 8, α-6, 3| = 0

3. Expand. Expanding along the first row gives (6+α-6) + (3α+8) + (α²-6α-16). This simplifies to α² - 2α - 8 = 0, that is (α-4)(α+2) = 0.
4. So. The values are α = 4 and α = -2. Their sum is 4 + (-2) = 2, which is option (d).

The idea to keep. For coplanarity of four points, fix one point, form three vectors from it and set the determinant to zero. When asked for a sum of values, the quadratic's -b/a gives it directly.

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