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

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Find the angle between the line

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

and the plane $3x + 4y + z + 5 = 0$

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:

The given line is $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{2-3}{2}$

\Rightarrow \overrightarrow{\mathrm{r}}=(2 \overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})+\mathrm{t}(3 \overrightarrow{\mathrm{i}} \overrightarrow{-\mathrm{j}}-2 \overrightarrow{\mathrm{k}})

$\therefore$ the vector parallel to the given line is

\vec{r}=3 \vec{i}-\vec{j}-2 \vec{k}

The given plane is $3 x+ \quad$ 4y+ $\quad z=-5$

\Rightarrow \vec{r}-(3 \vec{i}+4 \vec{j}-\vec{k})=5

$\therefore$ the normal to the given plane is

\overrightarrow{\vec{n}}=3 \vec{i}+4 \vec{j}+\vec{k}

Let $\theta$ be the angle between the line and the plane, then

$\begin{array}{l} \sin \theta=\frac{\vec{b} \cdot \vec{n}}{\vec{b}(\overrightarrow{n})} \\ =\frac{(3 \vec{i}-\vec{j}-2 \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+\vec{k})}{|3 \vec{i}-\vec{j}-2 \vec{k}|\left|3 \vec{i}\right| 4 \vec{j}+\vec{k} \mid} \\ =\frac{(3)(3)+(-1)(4)+(-2)(1)}{\sqrt{3^{2}+(-1)^{2}}+(-2)^{2} \sqrt{3^{2}+1^{2}+1^{2}}} \\ =\frac{9-4-2}{\sqrt{9+1+4 \sqrt{9+16+1}}}=\frac{3}{\sqrt{14 \sqrt{25}}}=\frac{3}{2 \sqrt{91}} \\ $\theta$=\sin ^{-1}\left(\frac{3}{2 \sqrt{91}}\right) \end{array}$