If $\overline{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\overline{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ are two vectors,then the angle between the vectors $3 \bar{a}+5 \bar{b}$ and $5 \bar{a}+3 \bar{b}$ is

$a, b$ and $c$ are three vectors such that $|a|=1, |b|=2, |c|=3$ and $b, c$ are perpendicular. If the projection of $b$ on $a$ is the same as the projection of $c$ on $a$, then $|a-b+c|$ is equal to

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ and $|\vec{a} - \vec{b}| = \sqrt{7}$,then find $|\vec{b}|$.

The projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$ is:

If $\overrightarrow{a} = \hat{i} + 2\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$,such that $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{c}$ is equal to: