Given $\vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k})$,$\vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k})$. The value of $(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

Among three vectors,two vectors are equal in magnitude,and the magnitude of the third vector is $\sqrt{2}$ times the magnitude of the other two vectors. If $\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0$,find the angles between the vectors.

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $x$-axis,respectively. The magnitudes of these vectors are $5 \ m$ and $12 \ m$,respectively. The angle that their resultant vector makes with the $x$-axis is:

$A$ vector of length $\ell$ is turned by an angle $\theta$. Find the change in the position vector of the tip.

The vector sum of two forces is perpendicular to their vector difference. In that case,the forces

The vectors $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ are perpendicular to each other. This is possible under the condition: