Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$,$\vec{a} \cdot \vec{b}=3$ and $|\vec{a} \times \vec{b}|^{2}=75$. Then $|\vec{a}|^{2}$ is equal to:

If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

The set of all real values of $c$ such that the angle between the vectors $\vec{a} = cx \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2cx \hat{k}$ is an obtuse angle for all real $x$ is:

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

Using vectors,prove that $\cos (A-B)=\cos A \cos B+\sin A \sin B$.

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is