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 $]$

There are two force vectors, one of $5\, N$ and other of $12\, N $ at what angle the two vectors be added to get resultant vector of $17\, N, 7\, N $ and $13 \,N$ respectively

The five sides of a regular pentagon are represented by vectors $A _1, A _2, A _3, A _4$ and $A _5$, in cyclic order as shown below. Corresponding vertices are represented by $B _1, B _2, B _3, B _4$ and $B _5$, drawn from the centre of the pentagon.Then, $B _2+ B _3+ B _4+ B _5$ is equal to

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

Write two properties of vector addition.

Two forces, ${F_1}$ and ${F_2}$ are acting on a body. One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is