If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b},$ then $\vec{a} \cdot \vec{b} \ge 0$ if

Find $|\vec{x}|$,if for a unit vector $\vec{a}$,$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$.

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

The angle between two vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

If $|a + b| > |a - b|$,then the angle between $a$ and $b$ is

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. $A$ vector of the type $\vec{b} + \lambda \vec{c}$ for some scalar $\lambda$,whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is