If the angle between two vectors $\vec{u} = i + k$ and $\vec{v} = i - j + ak$ is $\pi / 3,$ then the value of $a$ is:

If $\overrightarrow{a} = \hat{i} + 2\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$,such that $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{c}$ is equal to:

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ 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 $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

$A, B, C, D$ are any four points,then $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $