If $\overline{a}+\overline{b}+\overline{c}=\overline{0}$ with $|\overline{a}|=3, |\overline{b}|=5$ and $|\overline{c}|=7$,then the angle between $\overline{a}$ and $\overline{b}$ is:

If vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ and vector $\vec{b} = -2\hat{i} + 2\hat{j} - \hat{k},$ then $\frac{\text{Projection of vector } \vec{a} \text{ on vector } \vec{b}}{\text{Projection of vector } \vec{b} \text{ on vector } \vec{a}} = $

If $a \perp b$ and $(a+b) \perp (a+mb)$,then $m$ is equal to

If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$,$\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$,$\vec{c}=3\hat{i}+\hat{j}$ and $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then $\lambda=$

Let the position vectors of the vertices of a triangle $ABC$ be $\bar{a}, \bar{b}, \bar{c}$. If on the plane of the triangle,$P$ is a point having position vector $\bar{x}$ such that $\bar{x} \cdot (\bar{c} - \bar{b}) = \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b}$ and $\bar{x} \cdot (\bar{a} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{b} \cdot \bar{c}$,then for the triangle $ABC$,$P$ is the

If $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$ are orthogonal and $|\vec{a}|=|\vec{b}|$,then $(\lambda, \mu) = $