BM Sharma Solutions for Chapter: Vectors, Exercise 2: DPP

The resultant of $\overrightarrow{\mathit{A}}\mathit{+}\overrightarrow{\mathit{B}}$ is ${\overrightarrow{R}}_1$. On reversing the vector $\overrightarrow{\mathit{B}}$, the resultant becomes ${\overrightarrow{\mathit{R}}}_{\mathit{2}}\mathit{.}$ What is the value of $R_1^2+R_2^2?$

The resultant of $\overrightarrow{P}$ and $\overrightarrow{\mathit{Q}}$ is perpendicular to $\overrightarrow{\mathrm{P}}$. What is the angle between $\overrightarrow{P}$ and $\overrightarrow{\mathit{Q}}?$

Two forces $3 \mathrm{N}$ and $2 \mathrm{N}$ are at an angle $\text{θ}$ such that the resultant is $R$. The first force is now increased to $6 \mathrm{N}$ and the resultant become $2R$. The value of $\text{θ}$ is:

If $\left|\overrightarrow{\mathit{A}}\mathit{+}\overrightarrow{\mathit{B}}\right|=\left|\overrightarrow{A}\right|+\left|\overrightarrow{\mathit{B}}\right|$, then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ will be:

The maximum and minimum magnitude of the resultant of two vectors are $17$ units and $7$ units respectively. If these two vectors are at right angles to each other, the magnitude of their resultant is:

Two equal forces ($P$ each) act at a point inclined to each other at an angle of $120°$. The magnitude of their resultant is:

The sum of the magnitudes of two forces acting at a point is $18$ and the magnitude of their resultant is $12$. If the resultant is at $90°$ with the force of smaller magnitude, what are the magnitudes of forces?

Three forces $P$, $Q$ and $R$ are acting on a particle in the plane. The angle between $P$ and $Q$ is $150°$ and between $Q$ and $R$ is $120°$. Then, for equilibrium condition, the forces $P$, $Q$ and $R$ are in the ratio of