If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $15$ square units,then the area (in square units) of the parallelogram,having $3 \bar{a} + 2 \bar{b}$ and $\bar{a} + 3 \bar{b}$ as two adjacent sides,is

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

$i \times (j \times k) = $

If $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$, then $\vec{a} \cdot(\vec{c} \times \vec{b}-\vec{b}-\vec{c})=$

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors such that $\bar{a} \cdot \bar{b} = \frac{1}{2}$,$\bar{c} \cdot \bar{d} = \frac{1}{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$,then the value of $|[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}| = $

If $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$,and $a, b, c$ are the moduli of the vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then: