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Question: If the vectors \(2\mathbf{i} - \mathbf{j} + \mathbf{k},\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\) and ...

If the vectors 2i−j+k,i+2j−3k2\mathbf{i} - \mathbf{j} + \mathbf{k},\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} and 3i+λj+5k3\mathbf{i} + \lambda\mathbf{j} + 5\mathbf{k}

be coplanar, then λ=\lambda =

A

– 1

B

– 2

C

– 3

D

– 4

Answer

– 4

Explanation

Solution

If the given vectors are coplanar, then their scalar triple product is zero. ∣2−1112−33λ5∣=0⇒λ=−4.\left| \begin{matrix} 2 & - 1 & 1 \\ 1 & 2 & - 3 \\ 3 & \lambda & 5 \end{matrix} \right| = 0 \Rightarrow \lambda = - 4.