If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then $a \cdot b+b \cdot c+c \cdot a$ is equal to

$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$, then $\sin (\alpha + \beta) =$

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)

If the points $P$ and $Q$ are respectively the circumcentre and the orthocentre of a $\triangle ABC$,then $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}$ is equal to

If $a$ and $b$ are mutually perpendicular vectors,then $(a + b)^2 = $

If $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{k}$, then the point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is