$A$ 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:

If $|a| = 2$,$|b| = 5$ and $|a \times b| = 8$,then $a \cdot b$ is equal to

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$,$|\bar{a}|=3$,$|\bar{b}|=4$,and $|\bar{c}|=5$. Then,find the value of $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$.

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then:

Three vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$,then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.