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 $a \neq 0, b \neq 0, c \neq 0, a \times b = 0$ and $b \times c = 0$,then $a \times c$ is equal to

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$,with $|\lambda| \leq 5$,are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units,then $5-6 \lambda$ is equal to $.........$.

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$, then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

The unit vector perpendicular to the vectors $i - j + k$ and $2i + 3j - k$ is

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: