Consider three vectors $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$. $A$ vector $X$ of the form $X = \alpha A + \beta B$ (where $\alpha$ and $\beta$ are scalars) is perpendicular to $C$. The ratio of $\alpha$ to $\beta$ is: (in $: 1$)

$A$ vector perpendicular to the vector $(4 \hat{i}-3 \hat{j})$ is

Two unit vectors $\hat{a}_{1}$ and $\hat{a}_{2}$ are inclined to each other at an angle $\theta$. If $|\hat{a}_{1}-\hat{a}_{2}|=\sqrt{3}$,then the value of $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$ 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 ...... .

What is the angle between the resultant of $\overrightarrow{A} + \overrightarrow{B}$ and $\overrightarrow{A} \times \overrightarrow{B}$?

If $\overrightarrow A = 3\hat i + \hat j + 2\hat k$ and $\overrightarrow B = 2\hat i - 2\hat j + 4\hat k$,then the value of $|\overrightarrow A \times \overrightarrow B |$ will be: