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

If the vectors $\overline{\mathrm{a}}=\mathrm{c}\left(\log _7 x\right) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \quad$ and $\overline{\mathrm{b}}=\left(\log _\gamma x\right) \hat{\mathrm{i}}+3 \mathrm{c}\left(\log _\gamma x\right) \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ make obtuse angle for any $x>0$, then c belongs to

The altitude through vertex $A$ of $\triangle A B C$ with position vectors of points $A, B, C$ as $\bar{a}, \bar{b}, \bar{c}$ respectively is

If $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are unit vectors and $|\overline{\mathrm{a}}|=7$, $\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}} \times(\overline{\mathrm{c}} \times \overline{\mathrm{a}})=\frac{1}{2} \overline{\mathrm{a}}$, then angle between the vectors $\bar{a}$ and $\bar{c}$ and angle between the vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are respectively