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 $]$
$9.25 \hat{{i}}+5 \hat{{j}}$
$3 \hat{{i}}+15 \hat{{j}}$
$2.5 \hat{i}-14.5 \hat{{j}}$
$-1.5 \hat{{i}}-15.5 \hat{{j}}$
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $\mathrm{R}$ are inclined at an angle $\theta$, then
Two forces of $10 \,N$ and $6 \,N$ act upon a body. The direction of the forces are unknown. The resultant force on the body may be .........$N$
Three forces given by vectors $2 \hat{i}+2 \hat{j}, 2 \hat{i}-2 \hat{j}$ and $-4 \hat{i}$ are acting together on a point object at rest. The object moves along the direction
Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $‘n’$ times the magnitude of $(\vec A - \vec B)$. The angle between $ \vec A$ and $\vec B$ is