Let a, b and c be distinct positive numbers. If the vectors aî + aĵ + ck̂,î+k̂ and cî + cĵ + bk̂ are co-planar, then c is equal to :

Solution

Maths sir · 3 steps

Three vectors are coplanar exactly when their determinant is zero, so set it up and expand.

1. Coplanar condition. Three vectors lie in one plane exactly when their scalar triple product is zero. So write their components as the rows of a determinant and set it equal to zero.

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

2. Expand along row 1. a(0 · b - 1 · c) - a(1 · b - 1 · c) + c(1 · c - 0 · c) = -ac - ab + ac + c². The ac terms cancel, which leaves c² - ab = 0.
3. So. c² = ab. Since c is positive, c = √ab. That is option (d): c is the geometric mean of a and b.

The idea to keep. For coplanarity, put the three vectors as rows of a determinant and set it to zero. A result like c² = ab means a, c, b are in GP, so c is the geometric mean.

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