For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are

Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{|\vec{a} \times \vec{b}|}{2}$. Also,find the area of the parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}$.

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

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

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

The point of intersection of $r \times a = b \times a$ and $r \times b = a \times b$,where $a = i + j$ and $b = 2i - k$ is