91Ó°ÊÓ

To find the normal vector to a plane determined by two vectors, calculate the cross product of those vectors. For the given vectors, calculate: - Normal to \( \mathbf{i} + \mathbf{j} \) and \( \mathbf{j} + \mathbf{k} \): \( \mathbf{n}_1 = (\mathbf{i} + \mathbf{j}) \times (\mathbf{j} + \mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 0 & 1 & 1 \end{vmatrix} = \mathbf{i} - \mathbf{j} + \mathbf{k} \)- Normal to \( \mathbf{j} + \mathbf{k} \) and \( \mathbf{k} + \mathbf{i} \): \( \mathbf{n}_2 = (\mathbf{j} + \mathbf{k}) \times (\mathbf{k} + \mathbf{i}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & 1 \ 1 & 0 & 1 \end{vmatrix} = \mathbf{i} + \mathbf{j} - \mathbf{k} \)- Normal to \( \mathbf{k} + \mathbf{i} \) and \( \mathbf{i} + \mathbf{j} \): \( \mathbf{n}_3 = (\mathbf{k} + \mathbf{i}) \times (\mathbf{i} + \mathbf{j}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 1 \ 1 & 1 & 0 \end{vmatrix} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \)

Unit vectors are required, so normalize the normal vectors found in Step 1. Calculate the magnitude of each vector and divide each component by the magnitude:- \( |\mathbf{n}_1| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \), so \( \hat{\mathbf{n}}_1 = \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} + \mathbf{k}) \)- \( |\mathbf{n}_2| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \), so \( \hat{\mathbf{n}}_2 = \frac{1}{\sqrt{3}}(\mathbf{i} + \mathbf{j} - \mathbf{k}) \)- \( |\mathbf{n}_3| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \), so \( \hat{\mathbf{n}}_3 = \frac{1}{\sqrt{3}}(-\mathbf{i} + \mathbf{j} + \mathbf{k}) \)

The volume of the parallelepiped is given by the scalar triple product of the unit vectors. Compute:\[ V = \left( \hat{\mathbf{n}}_1 \times \hat{\mathbf{n}}_2 \right) \cdot \hat{\mathbf{n}}_3 \]Calculating the cross product first:\( \hat{\mathbf{n}}_1 \times \hat{\mathbf{n}}_2 = \frac{1}{3} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = \frac{1}{3}(0\mathbf{i} - 0\mathbf{j} - 2\mathbf{k}) = \frac{-2}{3}\mathbf{k} \)Then calculate the dot product with \( \hat{\mathbf{n}}_3 \):\( V = \left( \frac{-2}{3}\mathbf{k} \right) \cdot \left( \frac{1}{\sqrt{3}}(-\mathbf{i} + \mathbf{j} + \mathbf{k}) \right) = \frac{-2}{3\sqrt{3}} \)Since volume is positive, \( V = \frac{2}{3\sqrt{3}} \).

The calculated volume \( \frac{2}{3\sqrt{3}} \) needs to be simplified for comparison with options.Multiply numerator and denominator by \(\sqrt{3}\) to rationalize the denominator:\[ V = \frac{2}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{9} \]This does not match any given choice, indicating an error in previous steps. Verify and identify correct computation with correct expressions.