Consider three vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$. Let $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}=\overrightarrow{b} \times \overrightarrow{c}$. If $\alpha \in [0, \frac{\pi}{3}]$ is the angle between the vectors $\overrightarrow{b}$ and $\overrightarrow{c}$,then the minimum value of $27|\overrightarrow{c}-\overrightarrow{a}|^2$ is equal to :

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$. Then $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,$\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$,where $\alpha \in R$,then the value of $\alpha + |\overrightarrow{r}|^{2}$ is equal to:

If $|a| = 2$,$|b| = 5$ and $|a \times b| = 8$,then $a \cdot b$ is equal to

$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

Let $a$ and $b$ be two unit vectors inclined at an angle $\theta$,then $\sin(\theta/2)$ is equal to