If $a$ is a vector of magnitude $7$ and $b$ is a vector of magnitude $8$,then the maximum value of $|a \cdot b|$ is:

If $a=\hat{i}+\hat{j}+\hat{k}$, $a \cdot b=1$ and $a \times b=\hat{j}-\hat{k}$, then $b=$

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$,find the number of possible vectors $\vec{b}$ such that $\vec{a} \cdot \vec{b} = 10$,where $(x, y, z) \in \mathbb{N}$.

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \cdot(\overline{b}+\overline{c})+\overline{b} \cdot(\overline{c}+\overline{a})+\overline{c} \cdot(\overline{a}+\overline{b})=0$ and $|\overline{a}|=1$,$|\overline{b}|=8$ and $|\overline{c}|=4$,then $|\overline{a}+\overline{b}+\overline{c}|$ has the value

The magnitude of vectors $\vec{a}$ and $\vec{b}$ are $1$ and $2$ respectively,and $\vec{a} \cdot \vec{b} = 1$. Then,the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$,if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$,then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = $ . . . . . . .