Find the angle between the vectors $2\hat{i} + 3\hat{j} + \hat{k}$ and $2\hat{i} - \hat{j} - \hat{k}$.

The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

If the vectors $\overline{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\overline{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$ are mutually orthogonal,then $(\lambda, \mu) = $

If $a = 3i - j + 2k$ and $b = 2i + j - k$,then evaluate $a \times (a \cdot b)$.

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is