If $|a \times b|^2 + |a \cdot b|^2 = 144$ and $|a| = 4$,then $|b|$ is equal to

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

If $\theta$ is an obtuse angle between vectors $\overline{a}$ and $\overline{b}$ such that $|\overline{a}|=5$,$|\overline{b}|=3$ and $|\overline{a} \times \overline{b}|=5 \sqrt{5}$,then $\overline{a} \cdot \overline{b}=$

If $A=(0,4,-3)$,$B=(5,0,12)$,and $C=(7,24,0)$,then $\angle BAC=$

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

Two adjacent sides of a parallelogram $ABCD$ are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ makes a right angle with the side $\vec{AB}$, then the cosine of the angle $\alpha$ is...