Question

If vectors $\hat{p},\hat{q}and\hat{r}$ have magnitudes 5,12 and 13 units and $\hat{p}+\hat{q}=\hat{r}$, find the angle between $\hat{q}and\hat{r}$

Answer 1

Given that,

$|\overrightarrow{P}|=5$

$|\overrightarrow{Q}|=12$

$|\overrightarrow{R}|=13$

$\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$

We know that,

$\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$

$\overrightarrow{R}-\overrightarrow{Q}=\overrightarrow{P}$

Now,

$(\overrightarrow{R}-\overrightarrow{Q})\cdot (\overrightarrow{R}-\overrightarrow{Q})=\overrightarrow{P}\cdot \overrightarrow{P}$

$R^2-2RQ\cos \theta +Q^2=P^2$

$\cos \theta =\frac{R^2+Q^2-P^2}{2R\times Q}$

$\cos \theta =\frac{{13}^2+{12}^2-5^2}{2\times 13\times 12}$

$\cos \theta =\frac{12}{13}$

$\theta ={\cos }^{-1}\frac{12}{13}$

Hence, the angle between $\overrightarrow{Q}$ and$\overrightarrow{R}$ is ${\cos }^{-1}\frac{12}{13}$