If the angle between the vectors $a$ and $b$ is $\theta$ and $a \cdot b = \cos \theta$,then the true statement is:

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

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

The vector $b = 3j + 4k$ is to be written as the sum of a vector $b_1$ parallel to $a = i + j$ and a vector $b_2$ perpendicular to $a$. Then $b_1 = $

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,$|\overrightarrow{a}|=3$,$|\overrightarrow{b}|=5$,and $|\overrightarrow{c}|=7$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$