If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$,then $|\bar{b}|=$

Magnitudes of vectors $\vec a, \vec b, \vec c$ are $3, 4, 5$ respectively. If $\vec a$ and $\vec b + \vec c$,$\vec b$ and $\vec c + \vec a$,and $\vec c$ and $\vec a + \vec b$ are mutually perpendicular,then find the magnitude of $|\vec a + \vec b + \vec c|$.

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

If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

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

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?