Let a, b, c be three distinct real numbers, none equal to one. If the vectors a î+ĵ+k̂, î+b ĵ+k̂ and î+ĵ+c k̂ are coplanar, then 1/(1-a)+1/(1-b)+1/(1-c) is equal to :

Solution

Maths sir · 5 steps

Coplanar means the determinant is zero. Subtract rows so every entry turns into a 1 − a type term.

1. Coplanar condition. Three vectors are coplanar exactly when their scalar triple product is zero. Put their components in the rows of a determinant and set it equal to zero.

   |a, 1, 1; 1, b, 1; 1, 1, c| = |a, 1, 1; 1-a, b-1, 0; 1-a, 0, c-1| = 0

2. Row operations, expand. The second determinant comes from R₂ → R₂ - R₁ and R₃ → R₃ - R₁, which do not change its value. Expanding along the first row gives a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0.
3. Substitute. Let p = 1-a, q = 1-b, r = 1-c, so a = 1-p, b-1 = -q and c-1 = -r. The equation becomes (1-p)qr + pr + pq = 0, that is, qr - pqr + pr + pq = 0.
4. Divide by pqr. None of a, b, c is 1, so p, q, r are all non-zero and you can divide by pqr. This gives 1/p - 1 + 1/q + 1/r = 0.
5. So. 1/(1-a) + 1/(1-b) + 1/(1-c) = 1, which is option (b).

The idea to keep. When the answer is in terms of 1-a, 1-b, 1-c, make those terms appear in the determinant with row operations. Then substitute and divide by their product.

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