Question

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

• A. \( \dfrac{1}{3}( \hat{i} + \hat{j} + \hat{k} ) \)
• B. \( \dfrac{1}{3}( \hat{i} - \hat{j} + \hat{k} ) \)
• C. \( \dfrac{1}{3}( \hat{i} + \hat{j} - \hat{k} ) \)
• D. \( \dfrac{1}{3}( -\hat{i} - \hat{j} - \hat{k} ) \)

Hint:

The cross product of two vectors always gives a vector perpendicular to both of them. To obtain a unit vector, divide the resulting vector by its magnitude.

Answer 1

The Correct Option is A

Concept: A vector perpendicular to both vectors \( \vec{a} \) and \( \vec{b} \) is obtained using the cross product: \[ \vec{a} \times \vec{b} \] The unit vector in that direction is: \[ \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]
Step 1: {Find the cross product \( \vec{a} \times \vec{b} \).}
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 2 & -3 \end{vmatrix} \] \[ = \hat{i}((-1)(-3) - (1)(2)) - \hat{j}((2)(-3) - (1)(1)) + \hat{k}((2)(2) - (-1)(1)) \] \[ = \hat{i}(3 - 2) - \hat{j}(-6 - 1) + \hat{k}(4 + 1) \] \[ = \hat{i} + 7\hat{j} + 5\hat{k} \]
Step 2: {Find the magnitude of the vector.}
\[ |\vec{a} \times \vec{b}| = \sqrt{1^2 + 7^2 + 5^2}
= \sqrt{1 + 49 + 25}
= \sqrt{75}
= 5\sqrt{3} \]
Step 3: {Find the unit vector.}
\[ \text{Unit vector} = \frac{\hat{i} + 7\hat{j} + 5\hat{k}}{5\sqrt{3}} \] Thus, the unit vector perpendicular to both vectors is in the direction of \( \vec{a} \times \vec{b} \).