If vec{a}=hat{i}+hat{j}-hat{k}, vec{b}=hat{i} | vedclass

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(B) Given $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$

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$\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$

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(B) Given $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ and perpendicular to $\vec{a}$, $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$.The vector $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{c}$. Since $\vec{c}$ is in the plane of $\vec{a}$ and $\vec{b}$, the vector $\vec{d}$ must be perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.Thus, $\vec{d}$ is parallel to $\vec{a} 	imes \vec{b}$.Calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - 1) - \hat{j}(1 - (-1)) + \hat{k}(-1 - 1) = 0\hat{i} - 2\hat{j} - 2\hat{k} = -2(\hat{j} + \hat{k})$.The unit vector $\vec{d}$ is $\pm \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \pm \frac{-2(\hat{j} + \hat{k})}{\sqrt{(-2)^2 + (-2)^2}} = \pm \frac{-2(\hat{j} + \hat{k})}{\sqrt{8}} = \pm \frac{-2(\hat{j} + \hat{k})}{2\sqrt{2}} = \pm \frac{1}{\sqrt{2}}(\hat{j} + \hat{k})$.

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

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