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

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$,then $\vec{c} \cdot \vec{b}$ is equal to

Let $\theta$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{a} \cdot \vec{b}=4$ and $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$,then $\vec{a}+\vec{b}$ is:

Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

The vectors $3 \vec{a}-5 \vec{b}$ and $2 \vec{a}+\vec{b}$ are mutually perpendicular and the vectors $\vec{a}+4 \vec{b}$ and $-\vec{a}+\vec{b}$ are also mutually perpendicular. Then the angle between the vectors $\vec{a}$ and $\vec{b}$ is:

If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.