Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$,where $|\vec{a}|=4, |\vec{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2} + 4(\vec{a} \cdot \vec{b})^{2}$ is equal to

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$,then $|\vec{a} \cdot \vec{d}|$ is equal to:

If $\vec{u}$ and $\vec{v}$ are unit vectors and $\theta$ is the acute angle between them,then $2\vec{u} \times 3\vec{v}$ is a unit vector for

$(a - b) \times (a + b) = $

$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :