If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\sqrt{3} \vec{c}=\overrightarrow{0}$,then the angle between $\vec{a}$ and $\vec{b}$ is

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. Let $\overline{c}$ be a vector such that $|\bar{c}-\bar{a}|=3$ and $|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\overline{c}$ and $\overline{a} \times \overline{b}$ is $30^{\circ}$,then $\overline{a} \cdot \overline{c}$ is equal to

If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$

If $S$ is the circumcentre, $O$ is the orthocentre and $G$ is the centroid of a triangle $ABC$, then match the items of the List-$I$ with those of the items of List-$II$ given below.
| List-$I$ | List-$II$ |
| :--- | :--- |
| $(i)$ $\vec{SA} + \vec{SB} + \vec{SC}$ | $(A)$ $2\vec{OS}$ |
| (ii) $\vec{GA} + \vec{GB} + \vec{GC}$ | $(B)$ $\frac{2}{3}\vec{OS}$ |
| (iii) $\vec{OA} + \vec{OB} + \vec{OC}$ | $(C)$ $\vec{0}$ |
| (iv) $\vec{OG}$ | $(D)$ $\vec{SO}$ |
| | $(E)$ $\vec{OS}$ |

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 the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is: