Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Exercise 1

If $\overrightarrow{r}=\alpha \left(\overrightarrow{b}\times \overrightarrow{c}\right)+\beta \left(\overrightarrow{c}\times \overrightarrow{a}\right)+\gamma \left(\overrightarrow{a}\times \overrightarrow{b}\right)$ and $\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]=2$, then $a+\beta +\gamma$ is equal to

For any three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}; \left(\overrightarrow{b}\times \overrightarrow{c}\right)\times \left(\overrightarrow{c}\times \overrightarrow{a}\right)=$

If the resultant of two forces of magnitudes $P$ and $Q$ acting at a point at an angle of $60°$ is $\sqrt{13}Q$, then find $\frac{P}{Q}$.

The unit vector perpendicular to the vectors $6\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ and $3\hat{\mathrm{i}}-6\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ is

If $\overrightarrow{a}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $\overrightarrow{b}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$, then the angle $\theta$ between $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by

For any vector $\overrightarrow{a}$, the value of $\hat{i}\times \left(\overrightarrow{a}\times \hat{i}\right)+\hat{j}\times \left(\overrightarrow{a}\times \hat{j}\right)+\overrightarrow{k}\times \left(\overrightarrow{a}\times \hat{k}\right)$ is

If $\left|\overrightarrow{\mathrm{a}}\right|=3, \left|\overrightarrow{\mathrm{b}}\right|=2, \left|\overrightarrow{\mathrm{c}}\right|=1$, then the value of $|\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}·\overrightarrow{\mathrm{a}}|$ is (given that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$)

A point $I$ is in the centre of a circle inscribed in a triangle $ABC$, then the vector sum $\left|\overrightarrow{BC}\right|\overrightarrow{IA}+\left|\overrightarrow{CA}\right|\overrightarrow{IB}+\left|\overrightarrow{AB}\right|\overrightarrow{IC}$ is