Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is:

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

$P$ is the circumcentre of $\triangle ABC$. If the position vectors of $A, B, C$ and $P$ are $\bar{a}, \bar{b}, \bar{c}$ and $\frac{\bar{a}+\bar{b}+\bar{c}}{4}$ respectively,then the position vector of the orthocentre of this triangle is

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 1$,$|\vec{b}| = 2$,and $|\vec{c}| = 3$,then find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$,then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+2(|\vec{a}|+|\vec{b}|+|\vec{c}|)=$

If $\bar{a}$ and $\bar{b}$ are two 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