Let the points $A, B$ and $P$ be $(-2, 2, 4), (2, 6, 3)$ and $(1, 2, 1)$ respectively. The magnitude of the moment of the force represented by $\overrightarrow{AB}$ and acting at $A$ about $P$ is

If vectors satisfy the condition $|a - c| = |b - c|$,then $(b - a) \cdot \left( c - \frac{a + b}{2} \right)$ is equal to

If $|\vec{a}|=\sqrt{26}$,$|\vec{b}|=7$ and $|\vec{a} \times \vec{b}|=35$,then $\vec{a} \cdot \vec{b}$ is-

The projection vector of $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ on $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ is . . . . . . .

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then:

Let $ABCD$ be a parallelogram where $\overrightarrow{AB} = \overrightarrow{a}$,$\overrightarrow{AD} = \overrightarrow{b}$,$|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$,where $\overrightarrow{a} \cdot \overrightarrow{b} > 0$. Then the area of this parallelogram is (in square units):