If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a}+2\vec{b}$ and $5\vec{a}-4\vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is:

Let $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} \times \vec{c})+\vec{b}=0$ and $\vec{a}.\vec{c}=4$, then $|\vec{c}|^2$ is equal to 

If $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ are the unit vectors such that $(\vec{a} \times \vec{b}).(\vec{c} \times \vec{d})=1$ and $(\vec{a}.\vec{c})=\frac{1}{2}$, then 

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are unit vectors, then $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 $ does not exceed

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{a}.\vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{j}-\hat{k}$, then $\vec{b}$ is equal to