The value of $\frac{(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2|\vec{b}|^2}$ is

Two forces $\vec{F_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{F_2} = -\hat{i} + 2\hat{j} - \hat{k}$ act on a particle. The particle is displaced from point $P(4\hat{i} - 3\hat{j} + 2\hat{k})$ to point $Q(6\hat{i} + \hat{j} + 3\hat{k})$. The work done by the forces is ............. units.

Let $\overrightarrow{a} = 2\hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{b}$ and $\overrightarrow{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statements:
$(A)$ $|\overrightarrow{a} + \lambda \overrightarrow{c}| \geq |\overrightarrow{a}|$ for all $\lambda \in R$.
$(B)$ $\overrightarrow{a}$ and $\overrightarrow{c}$ are always parallel.

Given vertices $A(3, -1)$,$B(2, 3)$,and $C(5, 1)$,find $m \angle A$.

Let two non-collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves so that at any time $t$ the position vector $\overline{OP}$ (where $O$ is the origin) is given by $\hat{a} \cos t + \hat{b} \sin t$. When $P$ is farthest from origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$. Then,

Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is: