Question

Find a vector of magnitude 5 which is perpendicular to both the vectors \( 3\hat{i} - 2\hat{j} + \hat{k} \text{ and } 4\hat{i} + 3\hat{j} - 2\hat{k} \).

Hint:

Quick Tip: To find a vector perpendicular to two vectors, always use the cross product. After calculating the cross product, normalize the vector if you need a specific magnitude.

Answer 1

Let the two given vectors be: \[ \mathbf{a} = 3\hat{i} - 2\hat{j} + \hat{k}, \quad \mathbf{b} = 4\hat{i} + 3\hat{j} - 2\hat{k} \] To find a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), we use the cross product \( \mathbf{a} \times \mathbf{b} \). \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 4 & 3 & -2 \end{vmatrix} \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} -2 & 1 \\ 3 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 4 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\ 4 & 3 \end{vmatrix} \] \[ = \hat{i} [(-2)(-2) - (1)(3)] - \hat{j} [(3)(-2) - (1)(4)] + \hat{k} [(3)(3) - (-2)(4)] \] \[ = \hat{i} [4 - 3] - \hat{j} [-6 - 4] + \hat{k} [9 + 8] \] \[ = \hat{i} [1] - \hat{j} [-10] + \hat{k} [17] \] \[ = \hat{i} + 10\hat{j} + 17\hat{k} \] Thus, the cross product \( \mathbf{a} \times \mathbf{b} = \hat{i} + 10\hat{j} + 17\hat{k} \). Now, the magnitude of the cross product is: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{1^2 + 10^2 + 17^2} = \sqrt{1 + 100 + 289} = \sqrt{390} \] Let the unit vector in the direction of \( \mathbf{a} \times \mathbf{b} \) be: \[ \hat{n} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} = \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}} \] Finally, we need a vector of magnitude 5 in this direction. The required vector is: \[ \mathbf{v} = 5 \cdot \hat{n} = 5 \cdot \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}} = \frac{5(\hat{i} + 10\hat{j} + 17\hat{k})}{\sqrt{390}} \] Thus, the required vector is: \[ \mathbf{v} = \frac{5\hat{i} + 50\hat{j} + 85\hat{k}}{\sqrt{390}} \]