If $a$ makes an acute angle with $b$,$r \cdot a = 0$ and $r \times b = c \times b$,then $r=$

If $\vec{\lambda}$ is a unit vector perpendicular to the plane of vectors $\vec{a}$ and $\vec{b}$,and the angle between them is $\theta$,then $\vec{a} \cdot \vec{b}$ will be:

$A, B, C, D$ are four points in a plane with position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ respectively such that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$. Then the point $D$ is the $\dots$ of $\triangle ABC$.

If $a, b$ and $c$ are mutually perpendicular vectors of the same magnitude,then the cosine of the angle between $a$ and $a+b+c$ is

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.

If $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors and $\overline{p}=\frac{\overline{b} \times \overline{c}}{[\overline{a} \overline{b} \overline{c}]}, \overline{q}=\frac{\overline{c} \times \overline{a}}{[\overline{a} \overline{b} \overline{c}]}, \overline{r}=\frac{\overline{a} \times \overline{b}}{[\overline{a} \overline{b} \overline{c}]}, \quad$ then $2 \overline{a} \cdot \overline{p}+\overline{b} \cdot \overline{q}+\overline{c} \cdot \overline{r}=$