Let $x$ and $y$ be real numbers. If $\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$,then $|\vec{a} \times \vec{b}|$ is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$

If the area of a parallelogram,whose diagonals are $\vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \alpha\hat{k}$,is $\frac{\sqrt{93}}{2}$ sq. unit,then $\alpha = $

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b}),$ where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.$

The area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = 2\hat{j} + 3\hat{k}$ is

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is: