If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\overrightarrow{c} = 2\hat{i} + 3\hat{j} - \hat{k}$,and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i} + \hat{j} + \hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i} + \alpha\hat{j} + \hat{k}$,then the value of $\alpha$ is:

Vectors $\bar{a}$ and $\bar{b}$ are such that $|\bar{a}|=1$,$|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

The unit vector perpendicular to both the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ is

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?

If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them,then for what value of $\theta$ is $2\hat{u} \times 3\hat{v}$ a unit vector?