Question

If the vectors $ \mathbf{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}, \quad \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{c} = m\hat{i} - \hat{j} + 2\hat{k} $ are coplanar, then the value of $ m $ is

• A. \( \frac{5}{8} \)
• B. \( \frac{5}{3} \)
• C. \( -\frac{7}{4} \)
• D. \( \frac{3}{7} \)

Hint:

The scalar triple product can be used to determine if three vectors are coplanar. If the product equals zero, the vectors are coplanar.

Answer 1

The Correct Option is B

For three vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) to be coplanar, their scalar triple product must be zero. The scalar triple product is given by: \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \] First, calculate the cross product \( \mathbf{b} \times \mathbf{c} \): \[ \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{c} = m\hat{i} - \hat{j} + 2\hat{k} \] \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & 2 & -1 m & -1 & 2 \end{vmatrix} \] Now calculate the determinant: \[ \mathbf{b} \times \mathbf{c} = \hat{i} \begin{vmatrix} 2 & -1 -1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ m & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ m & -1 \end{vmatrix} \] \[ = \hat{i} \left( (2)(2) - (-1)(-1) \right) - \hat{j} \left( (1)(2) - (-1)(m) \right) + \hat{k} \left( (1)(-1) - (2)(m) \right) \] \[ = \hat{i} (4 - 1) - \hat{j} (2 + m) + \hat{k} (-1 - 2m) \] \[ = 3\hat{i} - (2 + m)\hat{j} + (-1 - 2m)\hat{k} \] Next, calculate the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \): \[ \mathbf{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \] \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (2)(3) + (-3)(-(2 + m)) + (4)(-1 - 2m) \] \[ = 6 + 3(2 + m) + 4(-1 - 2m) \] \[ = 6 + 6 + 3m - 4 - 8m \] \[ = 8 - 5m \] For coplanarity, the scalar triple product must be zero: \[ 8 - 5m = 0 \] \[ m = \frac{8}{5} \] Thus, the value of \( m \) is \( \frac{5}{3} \).