If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . . (in $/2$)

If $|\bar{a}|=2, |\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors,then the area of the triangle whose vertices are $0, \bar{a}+2\bar{b}, \bar{a}-2\bar{b}$ is

The projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is:

Forces of magnitudes $3$ and $2$ units acting in the directions $5\hat{i} + 3\hat{j} + 4\hat{k}$ and $3\hat{i} + 4\hat{j} - 5\hat{k}$ respectively act on a particle which is displaced from the points $(1, -1, -1)$ to $(3, 3, 1)$. The work done by the forces is equal to

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively, then $\frac{a \times b}{a \cdot b}=$

Show that the points $A, B$ and $C$ with position vectors $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ respectively form the vertices of a right-angled triangle.