If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $\mathrm{R}$ are inclined at an angle $\theta$, then
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{2} \mathrm{R} \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=2 \mathrm{R} \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=2 \mathrm{R} \cos \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=2 R \cos \left(\frac{\theta}{2}\right)$
The magnitude of vectors $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$ in the given figure are equal. The direction of $\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }$ with $x$-axis will be
What vector must be added to the two vectors $\hat i - 2\hat j + 2\hat k$ and $2\hat i + \hat j - \hat k,$ so that the resultant may be a unit vector along $X-$axis
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
Let the angle between two nonzero vectors $\overrightarrow A $ and $\overrightarrow B $ be $120^°$ and resultant be $\overrightarrow C $
A particle is situated at the origin of a coordinate system. The following forces begin to act on the particle simultaneously (Assuming particle is initially at rest)
${\vec F_1} = 5\hat i - 5\hat j + 5\hat k$ ${\vec F_2} = 2\hat i + 8\hat j + 6\hat k$
${\vec F_3} = - 6\hat i + 4\hat j - 7\hat k$ ${\vec F_4} = - \hat i - 3\hat j - 2\hat k$
Then the particle will move