Question

Find the unit vector perpendicular to both \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \).

• A. \( \dfrac{-\hat{i} + 7\hat{j} + 5\hat{k}}{5\sqrt{3}} \)
• B. \( \dfrac{\hat{i} - 7\hat{j} - 5\hat{k}}{5\sqrt{3}} \)
• C. \( \dfrac{-\hat{i} + 7\hat{j} + 5\hat{k}}{\sqrt{75}} \)
• D. \( \dfrac{\hat{i} + 7\hat{j} + 5\hat{k}}{5\sqrt{3}} \)

Hint:

The cross product \( \vec{a} \times \vec{b} \) always gives a vector perpendicular to both \( \vec{a} \) and \( \vec{b} \). To obtain a unit vector, divide the result by its magnitude.

Answer 1

The Correct Option is C

Concept: A vector perpendicular to both vectors \( \vec{a} \) and \( \vec{b} \) is given by their cross product: \[ \vec{a} \times \vec{b} \] To obtain the unit vector, divide the cross product by its magnitude. \[ \text{Unit vector} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]

Step 1: Compute the cross product \( \vec{a} \times \vec{b} \). \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & -1 & 1 1 & 2 & -3 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} -1 & 1 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 1 & 2 \end{vmatrix} \] \[ = \hat{i}(3-2) - \hat{j}(-6-1) + \hat{k}(4+1) \] \[ = \hat{i} + 7\hat{j} + 5\hat{k} \]

Step 2: Find the magnitude of the cross product. \[ |\vec{a} \times \vec{b}| = \sqrt{1^2 + 7^2 + 5^2} \] \[ = \sqrt{1 + 49 + 25} = \sqrt{75} \]

Step 3: Determine the unit vector. \[ \text{Unit vector} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \] \[ = \frac{\hat{i} + 7\hat{j} + 5\hat{k}}{\sqrt{75}} \] A unit vector perpendicular to both vectors can also be its negative. Thus, \[ \boxed{\dfrac{-\hat{i} + 7\hat{j} + 5\hat{k}}{\sqrt{75}}} \]