Let the angle between two nonzero vectors $\overrightarrow A $ and $\overrightarrow B $ be $120^°$ and resultant be $\overrightarrow C $
$\overrightarrow C $ must be equal to $|\overrightarrow A - \overrightarrow B |$
$\overrightarrow C $ must be greater than $|\overrightarrow A - \overrightarrow B |$
$\overrightarrow C $ must be less than $|\overrightarrow A - \overrightarrow B |$
$\overrightarrow C $ may be equal to $|\overrightarrow A - \overrightarrow B |$
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 these forces $\overrightarrow{O P}, \overrightarrow{O Q}, \overrightarrow{O R}, \overrightarrow{O S}$ and $\overrightarrow{{OT}}$ is approximately $\ldots \ldots {N}$.
[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$ Given $\hat{{i}}$ and $\hat{{j}}$ unit vectors along ${x}, {y}$ axis $]$
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
If $\vec A$ and $\vec B$ are two non-zero vectors such that $\left| {\vec A + \vec B} \right| = \frac{{\left| {\vec A - \vec B} \right|}}{2}$ and $\left| {\vec A} \right| = 2\left| {\vec B} \right|$ then the angle between $\vec A$ and $\vec B$ is