If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 2$, $|\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 4$, then $|\vec{a} - \vec{b}| = . . . . . . $.

Given $\vec{a}=3 \hat{i}-\hat{j}$,$\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$ and $\vec{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$,where $\overrightarrow{b_1}$ is parallel to $\vec{a}$ and $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,then $\overrightarrow{b_2}$ is equal to

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$, $\vec{b}$ and $\vec{c}$ is $\beta$, and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

The constant value $(\lambda + \mu)$ for which the lines $\vec{r} = (2\hat{i} + \hat{j} + \hat{k}) + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = (\hat{i} + \hat{j} - 3\hat{k}) + \mu(\hat{j} + 2\hat{k})$ intersect each other is equal to (where $\lambda$ and $\mu$ are parameters).

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

Let $\bar{a} = \bar{i} + 2\bar{j} + 3\bar{k}$, $\bar{b} = 2\bar{i} - 3\bar{j} + \bar{k}$, and $\bar{c} = 3\bar{i} + \bar{j} - 2\bar{k}$ be three vectors. If $\bar{r}$ is a vector such that $\bar{r} \cdot \bar{a} = 0$, $\bar{r} \cdot \bar{b} = -2$, and $\bar{r} \cdot \bar{c} = 6$, then find the value of $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k})$.