The angle between vector $\vec{Q}$ and the resultant of $(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})$ and $(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})$ is:
$0^{\circ}$
$\tan ^{-1} \frac{(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})}{2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}}}$
$\tan ^{-1}\left(\frac{P}{Q}\right)$
$\tan ^{-1}\left(\frac{2 Q}{P}\right)$
Unit vector parallel to the resultant of vectors $\vec A = 4\hat i - 3\hat j$and $\vec B = 8\hat i + 8\hat j$ will be
The vectors $\vec{A}$ and $\vec{B}$ are such that
$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
The angle between the two vectors is
The resultant of $\overrightarrow A + \overrightarrow B $ is ${\overrightarrow R _1}.$ On reversing the vector $\overrightarrow {B,} $ the resultant becomes ${\overrightarrow R _2}.$ What is the value of $R_1^2 + R_2^2$
A person goes $10\, km$ north and $20\, km$ east. What will be displacement from initial point........$km$
If $\overrightarrow A = 4\hat i - 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$ then magnitude and direction of $\overrightarrow A \, + \overrightarrow B $ will be