Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $1$ and $2$ respectively,given that $\vec{a} \cdot \vec{b} = 1$.

The points represented by $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar and $(\sin A)\vec{a} + (2\sin 2B)\vec{b} + (3\sin 3C)\vec{c} - 4\vec{d} = \vec{0}$. Then the least value of $\frac{21}{8}(\sin^2 A + \sin^2 2B + \sin^2 3C)$ is:

$\bar{a}, \bar{b}, \bar{c}$ are three non-coplanar and mutually perpendicular vectors of same magnitude $K$. If $\bar{r}$ is any vector satisfying $\bar{a} \times ((\bar{r}-\bar{b}) \times \bar{a}) + \bar{b} \times ((\bar{r}-\bar{c}) \times \bar{b}) + \bar{c} \times ((\bar{r}-\bar{a}) \times \bar{c}) = \bar{0}$,then $\bar{r} =$

If $\bar{a}$ and $\bar{b}$ are not perpendicular to each other,$\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \cdot \bar{c} = 0$,then $\bar{r} =$

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $c$ is a vector perpendicular to $b$,then $\left\{\frac{a \cdot(b \times c)}{|b \times c|^2}\right\}(b \times c)+\left\{\frac{a \cdot b}{|b|^2}\right\} b+\left\{\frac{a \cdot c}{|c|^2}\right\} c$ is equal to:

If the vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $|\bar{a}-\bar{c}|=|\bar{b}-\bar{c}|$,then $(\bar{b}-\bar{a}) \cdot \left(\bar{c}-\frac{\bar{a}+\bar{b}}{2}\right) = $