The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.

If $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$,then the value of $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times 2\vec{b})]$ is:

Let $\vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-\alpha \hat{j}+\hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $8 \sqrt{3}$ square units,then $\vec{a} \cdot \vec{b}$ is equal to ....... .

If the direction ratios of two lines $L_1$ and $L_2$ are given by $(1, -2, 2)$ and $(-2, 3, -6)$ respectively,then the direction ratios of the line which is perpendicular to the lines $L_1$ and $L_2$ are

The sine of the angle between the two vectors $3i + 2j - k$ and $12i + 5j - 5k$ will be

$A$ unit vector perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is: