\( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
We need to determine the direction cosines of a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
Step 1: Compute the cross product \( \mathbf{a} \times \mathbf{b} \). This vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Given: \[ \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \quad \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k} \] The cross product is calculated as: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \] \[ = \hat{i}((2)(1) - (3)(-1)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-1) - (2)(2)) \] \[ = \hat{i}(2 + 3) - \hat{j}(1 - 6) + \hat{k}(-1 - 4) \] \[ = 5\hat{i} + 5\hat{j} - 5\hat{k} \] Therefore, the vector perpendicular to \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \times \mathbf{b} = 5\hat{i} + 5\hat{j} - 5\hat{k} \).
Step 2: Calculate the magnitude of \( \mathbf{a} \times \mathbf{b} \). The magnitude is: \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{5^2 + 5^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]
Step 3: Determine the direction cosines. The direction cosines are the components of the unit vector in the direction of \( \mathbf{a} \times \mathbf{b} \). The unit vector \( \hat{u} \) is: \[ \hat{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \] The direction cosines are: \[ \cos \alpha = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad \cos \beta = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad \cos \gamma = \frac{-5}{5\sqrt{3}} = \frac{-1}{\sqrt{3}} \] The direction cosines of the vector perpendicular to \( \mathbf{a} \) and \( \mathbf{b} \) are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \).