Let a⃗, b⃗ and c⃗ be three non-zero non-coplanar vectors. Let the position vectors of four points A,B,C and D be a⃗ - b⃗ + c⃗ ,λ a⃗ - 3b⃗ + 4c⃗ , - a⃗ + 2b⃗ - 3c⃗ and 2a⃗ - 4b⃗ + 6c⃗ respectively. If AB⃗ ,AC⃗ and AD⃗ are coplanar, then λ is equal to __________.

Solution

Maths sir · 4 steps

Write the three vectors in a, b, c components and set their 3 by 3 determinant to zero.

1. Build the vectors. Subtract the position vector of A from each point. This gives AB⃗ = (λ - 1)a⃗ - 2b⃗ + 3c⃗, AC⃗ = -2a⃗ + 3b⃗ - 4c⃗ and AD⃗ = a⃗ - 3b⃗ + 5c⃗.
2. Coplanar condition. Because a⃗, b⃗, c⃗ are non-coplanar, the three vectors are coplanar exactly when the determinant of their coefficients is zero.

   |λ - 1, -2, 3; -2, 3, -4; 1, -3, 5| = 0

3. Expand. Expand along the first row: (λ - 1)(15 - 12) + 2(-10 + 4) + 3(6 - 3) = 0. That is 3λ - 3 - 12 + 9 = 0, so 3λ - 6 = 0.
4. So. λ = **2**.

The common slip: -6. Taking the middle cofactor term as -2(-6) with a plus sign in front, instead of -(-2)(-6), gives 3λ + 18 = 0 and λ = -6.

The idea to keep. When vectors are written in terms of three non-coplanar vectors, treat the coefficients like components. Coplanar means the coefficient determinant is zero.

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