For $\vec{a}$ and $\vec{b}$,$|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between $\vec{a}$ and $\vec{b}$ is . . . . . . .

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

Let $\overline{a}, \overline{b}, \overline{c}$ be three vectors such that $|\overline{a}|=\sqrt{3}$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$ and the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$. If $\overline{a}$ is perpendicular to the vector $\overline{b} \times \overline{c}$,then $|\overline{a} \times(\overline{b} \times \overline{c})|$ is equal to

If $2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$,then $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is equal to

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?

Find all vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2 \hat{j}+\hat{k}$ and $-\hat{i}+3 \hat{j}+4 \hat{k}$.