Let the vectors u⃗₁=î+ĵ+a k̂, u⃗₂=î+b ĵ+k̂ and u⃗₃=c î+ĵ+k̂ be coplanar. If the vectors v⃗₁=(a+b) î+c ĵ+c k̂, v⃗₂=a î+(b+c) ĵ+a k̂ and v⃗₃=b î+b ĵ+(c+a) k̂ are also coplanar, then 6(a+b+c) is equal to :

Solution

Maths sir · 4 steps

The second determinant collapses to 4abc, and that one fact unlocks the first condition.

1. Coplanar means zero. Three vectors are coplanar exactly when their scalar triple product is zero. So the determinant of their components must be zero.
2. First set. For u⃗₁, u⃗₂, u⃗₃, expand along the first row: 1(b-1) - 1(1-c) + a(1-bc) = 0. This gives a + b + c - abc - 2 = 0.
3. Second set. Expand the second determinant along its first row. The first minor is (b+c)(c+a) - ab = c(a+b+c), the second is a(a+c-b), and the third is b(a-b-c). Collecting terms, everything cancels except 4ab, multiplied by c.

   |a+b, c, c; a, b+c, a; b, b, c+a| = c[(a+b)(a+b+c) - a(a+c-b) + b(a-b-c)] = 4abc

4. So. Coplanarity of the v⃗ vectors gives 4abc = 0, so abc = 0. Put this in the first condition: a + b + c = 2, so 6(a+b+c) = 12, option (a).

The idea to keep. Coplanar vectors mean the scalar triple product, a 3 by 3 determinant, is zero. When a determinant is full of sums like a+b, expand it fully: it often collapses to a neat product like 4abc.

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