Let $\bar{a}$ and $\bar{b}$ be two non-collinear unit vectors. If $\bar{u}=\bar{a}-(\bar{a} \cdot \bar{b}) \bar{b}$ and $\bar{v}=\bar{a} \times \bar{b}$,then $|\bar{v}|=$

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $|\overline{a}+\overline{b}+\overline{c}|=1$,$\overline{c}=\lambda(\overline{a} \times \overline{b})$ and $|\overline{a}|=\frac{1}{\sqrt{3}}, |\overline{b}|=\frac{1}{\sqrt{2}}, |\overline{c}|=\frac{1}{\sqrt{6}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

If $\bar{a}$ and $\bar{b}$ are unit vectors such that $(\bar{a} + 2\bar{b})$ and $(5\bar{a} - 4\bar{b})$ are perpendicular to each other,then the angle between $\bar{a}$ and $\bar{b}$ is .....$^o$.

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors with magnitudes $2, 3$ and $4$ respectively, then the best upper bound of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ among the given values is

If $a=\hat{i}+\hat{j}+\hat{k}$, $a \cdot b=1$ and $a \times b=\hat{j}-\hat{k}$, then $b=$

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^{2}$ is equal to: