Question

Find the angle between the line $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{-2}$ and the plane $3x+4y+z+5=0$

Answer 1

The given line is $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{2-3}{-2}$
$\Rightarrow \overrightarrow{r}=(2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})+t(3\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k})$
$\therefore$ the vector parallel to the given line is
$\overrightarrow{r}=3\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k}$
The given plane is $3x+4y+z=-5$
$\Rightarrow \overrightarrow{r}=(3\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{k})=5$
$\therefore$ the normal to the given plane is
$\overrightarrow{n}=3\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{k}$
Let $\theta$ be the angle between the line and the plane, then
$\sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}$
$=\frac{(3\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k}).(3\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{k})}{|3\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k}||3\overrightarrow{i}+4\overrightarrow{j}+\overrightarrow{k}|}$
$=\frac{\left(3\right)\left(3\right)+(-1)\left(4\right)+(-2)\left(1\right)}{\sqrt{3^2+(-1{)}^2+(-2{)}^2}\sqrt{3^2+4^2+1^2}}$
$=\frac{9-4-2}{\sqrt{9+1+4}\sqrt{9+16+1}}=\frac{3}{\sqrt{14}\sqrt{26}}=\frac{3}{2\sqrt{91}}$
$\theta ={\sin }^{-1}\left(\frac{3}{2\sqrt{91}}\right)$