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

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$,$\vec{a} \cdot \vec{b}=3$ and $|\vec{a} \times \vec{b}|^{2}=75$. Then $|\vec{a}|^{2}$ is equal to:

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

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,

In a parallelogram $ABCD$,$|\overline{AB}| = a$,$|\overline{AD}| = b$ and $|\overline{AC}| = c$,then the value of $\overline{DA} \cdot \overline{AB}$ is:

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is: