If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then find $\vec{a} \cdot \vec{b}$.

Let the vectors $\overline{a}, \overline{b}, \overline{c}$ be such that $|\overline{a}|=2, |\overline{b}|=4$ and $|\overline{c}|=4$. If the projection of $\overline{b}$ on $\overline{a}$ is equal to the projection of $\overline{c}$ on $\overline{a}$ and $\overline{b}$ is perpendicular to $\overline{c}$,then the value of $|\overline{a}+\overline{b}-\overline{c}|$ is equal to

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 $|\vec{a}|=5, |\vec{b}|=13$ and $|\vec{a} \times \vec{b}|=25$. If $\frac{\pi}{2} < \theta < \pi$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then the value of $\vec{a} \cdot \vec{b}$ is:

If $\theta$ is an obtuse angle between vectors $\overline{a}$ and $\overline{b}$ such that $|\overline{a}|=5$,$|\overline{b}|=3$ and $|\overline{a} \times \overline{b}|=5 \sqrt{5}$,then $\overline{a} \cdot \overline{b}=$

If $a, b, c$ are three vectors such that $a = b + c$ and the angle between $b$ and $c$ is $\pi / 2$,then: