Question

Find the angle between $\overrightarrow{r}=(2\hat{i}-\hat{j})+\lambda (\hat{i}+\hat{j}+\hat{k})$ and $\overrightarrow{P}\cdot \overrightarrow{r}(2\hat{i}+3\hat{j}-4\hat{k})=3$.

Answer 1

Given,

$\begin{array}{c}\overline{r}=(2\hat{i}-\hat{j})+\lambda (\hat{i}+\hat{j}+\hat{k})\\ and,\\ \overline{p}.\overline{r}(2\hat{i}+3\hat{j}-4\hat{k})=3\end{array}$

then,point on a line (2,-1,0)

Dr's of the line (1,1,1).

Plane - $2\hat{i}+3\hat{j}-4\hat{k}=3$

Dr's of plane (2,3,-4)

So,$\begin{array}{c}\sin \theta =\frac{2+3-4}{\sqrt{+1+1}\sqrt{4+9+16}}=\frac{1}{\sqrt{3}\sqrt{29}}\\ \therefore \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{87}}\right)\end{array}$