D. C. Pandey Solutions for Chapter: Vectors, Exercise 1: Excercise 1

An electric dipole moment, $\overrightarrow{p}=\left(-\hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)\times {10}^{-29} \mathrm{C} \mathrm{m}$ is at the origin $\left(0, 0, 0\right)$ . The electric field due to this dipole at $\overrightarrow{r}=\left(+\hat{\mathrm{i}}+3\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right) \mathrm{m}$ (note that $\overrightarrow{r}·\overrightarrow{p}=0$) is parallel to:

Two forces $P$ and $Q$ of magnitude $2F$ and $3F$, respectively, are at an angle $\mathrm{\theta }$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\mathrm{\theta }$ is

Two vectors $A$ and $B$ have equal magnitudes. The magnitude of $\left(A+B\right)$ is $n$ times the magnitude of $\left(A-B\right)$. The angle between $A$ and $B$ is

If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $\left(A+B\right)$ and $\left(A-B\right)$, then the angle between two vectors is

The resultant $\overrightarrow{R}$ of $\overrightarrow{P}$ and $\overrightarrow{Q}$ is perpendicular to $\overrightarrow{P}$. Also $\left|\overrightarrow{P}\right|=\left|\overrightarrow{R}\right|$. The angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ is: $\left[\tan \left(45°\right)=1\right]$

$\overrightarrow{P}$ and $\overrightarrow{Q}$ are two non-zero vectors inclined to each other at an angle $\mathrm{\theta }, \hat{p}$ and $\hat{q}$ are unit vectors along $\overrightarrow{P}$ and $\overrightarrow{Q}$ respectively. The component of $\overrightarrow{P}$ in the direction of $\overrightarrow{Q}$ will be

The vectors $\left(\overrightarrow{A}+\overrightarrow{B}\right)$ and $\left(\overrightarrow{A}-\overrightarrow{B}\right)$ are at right angles to each other. This is possible under the condition