The vector component of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the direction of vector $\vec{b} = (\hat{i} + \hat{j})$ is:

Three particles $P, Q$ and $R$ are moving along the vectors $\vec{A}=\hat{i}+\hat{j}, \vec{B}=\hat{j}+\hat{k}$ and $\vec{C}=-\hat{i}+\hat{j}$ respectively. They strike a point and start to move in different directions. Now,particle $P$ is moving normal to the plane containing vectors $\vec{A}$ and $\vec{B}$. Similarly,particle $Q$ is moving normal to the plane containing vectors $\vec{A}$ and $\vec{C}$. The angle between the directions of motion of $P$ and $Q$ is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$. Then the value of $x$ is ...... .

If $|\overrightarrow{A}|=4$ unit,$|\overrightarrow{A}+\overrightarrow{B}|=10$ unit and $\overrightarrow{A} \cdot(\overrightarrow{A}+\overrightarrow{B})=20$ unit,then $|\overrightarrow{B}|=$ ?

If the component of the vector $\vec{A}$ along the vector $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$,then the ratio of magnitudes of vectors $\vec{A}$ and $\vec{B}$ is

For three vectors $\vec{A} = (-x \hat{i} - 6 \hat{j} - 2 \hat{k})$,$\vec{B} = (-\hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (-8 \hat{i} - \hat{j} + 3 \hat{k})$,if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$,then the value of $x$ is:

Show that $\vec{a} \cdot(\vec{b} \times \vec{c})$ is equal in magnitude to the volume of the parallelepiped formed by the three vectors $\vec{a}, \vec{b}$ and $\vec{c}$.