Question

Find the angle between the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z+2}{4}$ and the plane

$2x+y-3z+4=0$

Answer 1

Let the $\theta$ is the angle between the line and normal to the plane .

The equation of the line

$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z+2}{4}$

Now,

Converting the equation in the normal form

we have,

$\overrightarrow{r}=\left(\hat{i}-\hat{j}-2\hat{k}\right)+\lambda \left(3\hat{i}+2\hat{j}+4\hat{k}\right)$

And,

$\overrightarrow{r.}\left(2\hat{i}+\hat{j}-3\hat{k}\right)=-4$

Here $\overrightarrow{b}=3\hat{i}+2\hat{j}+4\hat{k}$ and $\overrightarrow{n}=2\hat{i}+\hat{j}-3\hat{k}$

Now,

$\begin{array}{c}\sin \theta =\left|\frac{\left(3\hat{i}+2\hat{j}+4\hat{k}\right)\cdot \left(2\hat{i}+\hat{j}-3\hat{k}\right)}{\sqrt{3^2+2^2+4^2}\sqrt{2^2+1^2+{\left(-3\right)}^2}}\right|\\ \\ \sin \theta =\left|\frac{6+2-12}{\sqrt{29}\sqrt{14}}\right|\\ \\ \sin \theta =\left|\frac{-4}{\sqrt{406}}\right|=\frac{4}{\sqrt{406}}\\ \\ \theta ={\sin }^{-1}\left(\frac{4}{\sqrt{406}}\right)\end{array}$