The angle between the vectors $2 \hat{k} - 3 \hat{j}$ and $\hat{i} - 2 \hat{k}$ is

$A$ unit vector in the $xy$-plane that makes an angle of $45^{\circ}$ with the vector $(i + j)$ and an angle of $60^{\circ}$ with the vector $(3i - 4j)$ is:

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

Three non-coplanar vectors $\bar{a}, \bar{b}, \bar{c}$ are the coterminous edges of a parallelepiped. If $\bar{a}$ and $\bar{b}$ determine the base of the parallelepiped, then its height is:

Let $u, v$ and $w$ be three vectors such that $u+v+w=0$,$|u|=3$,$|v|=5$ and $|w|=7$. Then the angle between $u$ and $v$ is (in $^{\circ}$)