The area of the parallelogram whose diagonals are $\frac{3}{2}i + \frac{1}{2}j - k$ and $2i - 6j + 8k$ is

If $a=\hat{i}+\hat{j}+\hat{k}$,$c=\hat{j}-\hat{k}$,$a \times b=c$ and $a \cdot b=3$,then $b$ is equal to

Let $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$,where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2} + 4)}$,then the value of $2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to

The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{c}=\hat{j}-\hat{k}$,$\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$,then $\vec{b}$ is equal to:

Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to