Question

The vectors \(3\hat{i}-\hat{j}+2\hat{k},2\hat{i}+\hat{j}+3\hat{k}\) and \(\hat{i}+λ\hat{j}-\hat{k}\) are coplanar if λ is :

• A. -2
• B. 0
• C. 2
• D. any real number

Answer 1

The Correct Option is A

To determine the value of \( \lambda \) that makes the vectors \(\mathbf{a} = 3\hat{i}-\hat{j}+2\hat{k}\), \(\mathbf{b} = 2\hat{i}+\hat{j}+3\hat{k}\), and \(\mathbf{c} = \hat{i}+\lambda\hat{j}-\hat{k}\) coplanar, we need to use the condition for coplanarity of vectors: the scalar triple product of these vectors must be zero.

The scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\) can be calculated using the determinant of a \(3 \times 3\) matrix whose rows are the components of the vectors:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & \lambda & -1 \end{vmatrix}$$

Expanding the determinant:

$$= 3 \begin{vmatrix} 1 & 3 \\ \lambda & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & 1 \\ 1 & \lambda \end{vmatrix}$$

Calculating each of the \(2 \times 2\) determinants:

\(\begin{vmatrix} 1 & 3 \\ \lambda & -1 \end{vmatrix} = 1(-1) - 3\lambda = -1 - 3\lambda\)

\(\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = 2(-1) - 3(1) = -2 - 3 = -5\)

\(\begin{vmatrix} 2 & 1 \\ 1 & \lambda \end{vmatrix} = 2\lambda - 1\)

Substituting these back into the expression:

$$= 3(-1 - 3\lambda) + 1(-5) + 2(2\lambda - 1)$$

$$= -3 - 9\lambda - 5 + 4\lambda - 2$$

Combine like terms:

$$= -10 - 5\lambda$$

For the vectors to be coplanar, set this equal to zero:

$$-10 - 5\lambda = 0$$

Solving for \( \lambda \):

$$-5\lambda = 10$$

$$\lambda = -2$$

Thus, the value of \( \lambda \) that makes the vectors coplanar is:

-2