Question

If the vectors $2\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$, $-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$, and $p\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ are coplanar, then the unit vector in the direction of the vector $9p\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$ is

• A. $\frac{1}{6}(2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})$
• B. $\frac{1}{\sqrt{57}}(5\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})$
• C. $\frac{1}{\sqrt{68}}(6\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})$
• D. $\frac{1}{9}(-7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k})$

Hint:

For coplanar vectors, set the scalar triple product determinant to zero to find unknowns. The unit vector is $\frac{\mathbf{v}}{|\mathbf{v}|}$.

Answer 1

The Correct Option is D

For vectors $\mathbf{u} = 2\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$, $\mathbf{v} = -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$, and $\mathbf{w} = p\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ to be coplanar, their scalar triple product must be zero: \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} 2 & 4 & -3 \\ -1 & 2 & 3 \\ p & -2 & 1 \end{vmatrix} = 0 \] Compute the determinant: \[ 2 (2 \cdot 1 - 3 \cdot (-2)) - 4 ((-1) \cdot 1 - 3 \cdot p) + (-3) ((-1) \cdot (-2) - 2 \cdot p) = 0 \] \[ 2 (2 + 6) - 4 (-1 - 3p) - 3 (2 - 2p) = 16 + 4 + 12p - 6 + 6p = 18p + 14 = 0 \] \[ 18p = -14 \implies p = -\frac{7}{9} \] The vector is: \[ 9p\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} = 9 \left(-\frac{7}{9}\right) \mathbf{i} - 4\mathbf{j} + 4\mathbf{k} = -7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} \] Magnitude: \[ |-7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}| = \sqrt{(-7)^2 + (-4)^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9 \] Unit vector: \[ \frac{-7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}}{9} = \frac{1}{9} (-7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) \] Option (4) is correct. Options (1), (2), and (3) do not match the computed unit vector.