The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ is

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle,passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

The position vectors of the points $A, B$ and $C$ are $i + j, j + k$ and $k + i$ respectively. The vector area of the $\Delta ABC = \pm \frac{1}{2} \vec{\alpha}$ where $\vec{\alpha} = $

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $30^{\circ}$,then the value of $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is equal to