$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$, then $\sin (\alpha + \beta) =$

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot (2 \vec{a}+7 \vec{b})$.

$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane, and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}|$ equals?

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,$\overrightarrow{OC} = 2\vec{a} + 3\vec{b}$,and $\overrightarrow{OD} = \vec{a} - 2\vec{b}$. If $|\vec{a}| = 3|\vec{b}|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a} \cdot \vec{b} = 0$. For some $x, y \in \mathbb{R}$,let $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})$. If $|\vec{c}| = 2$ and the vector $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$,then the value of $8 \cos^2 \alpha$ is . . . . .