Let $a, b, c$ be the position vectors of the vertices of a triangle $ABC$. The vector area of triangle $ABC$ is

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$,then $\cot^2 \theta=$

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{a}$ is equal to ...........

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

Let $ABC$ be a triangle and $P$ be a point inside $ABC$ such that $\overrightarrow{PA} + 2\overrightarrow{PB} + 3\overrightarrow{PC} = \vec{0}$. The ratio of the area of $\triangle ABC$ to that of $\triangle APC$ is

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d}=\vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$,then $|3 \lambda \hat{d}+\mu \vec{c}|^2$ is equal to . . . . . . .