Direction cosines of a vector perpendicular to $ \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} $ and $ \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} $ are:
Show Hint
- \( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \)
- \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
- \( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \)
\( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
The Correct Option is A
Solution and Explanation
We are asked to find the direction cosines of a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
Step 1: Find the cross product of \( \mathbf{a} \) and \( \mathbf{b} \) To find the direction cosines of a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), we first compute the cross product \( \mathbf{a} \times \mathbf{b} \). Given: \[ \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \quad \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} \] The cross product formula is: \[ \mathbf{a} \times \mathbf{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\1 & 2 & 3 \\2 & -1 & 1 \end{matrix} \right| \] Using the determinant formula for the cross product: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \left( \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \right) \] \[ = \hat{i} \left( (2)(1) - (3)(-1) \right) - \hat{j} \left( (1)(1) - (3)(2) \right) + \hat{k} \left( (1)(-1) - (2)(2) \right) \] \[ = \hat{i} \left( 2 + 3 \right) - \hat{j} \left( 1 - 6 \right) + \hat{k} \left( -1 - 4 \right) \] \[ = \hat{i}(5) - \hat{j}(-5) + \hat{k}(-5) \] \[ \mathbf{a} \times \mathbf{b} = 5\hat{i} + 5\hat{j} - 5\hat{k} \] Thus, the vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \times \mathbf{b} = 5\hat{i} + 5\hat{j} - 5\hat{k} \).
Step 2: Find the magnitude of \( \mathbf{a} \times \mathbf{b} \) The magnitude of the vector \( \mathbf{a} \times \mathbf{b} \) is: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(5)^2 + (5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]
Step 3: Find the direction cosines The direction cosines of a vector are given by the components of the unit vector in the direction of \( \mathbf{a} \times \mathbf{b} \). The unit vector is: \[ \hat{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \] Thus, the direction cosines are: \[ \cos \alpha = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad \cos \beta = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad \cos \gamma = \frac{-5}{5\sqrt{3}} = \frac{-1}{\sqrt{3}} \] Hence, the direction cosines of the vector perpendicular to \( \mathbf{a} \) and \( \mathbf{b} \) are \( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \).
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