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 . . . . . . .

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$,then $|\vec{x}| = $ . . . . . . .

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{3}$. If a parallelogram is constructed with adjacent sides $2\vec{a}+3\vec{b}$ and $\vec{a}-\vec{b}$,then its shorter diagonal is of length

Let $P$ be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\overrightarrow{AP}|=\frac{\sqrt{5}}{2}$, then the area of the triangle $ABP$ is:

Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ is:

If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,$\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$