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

$$\begin{gathered} \text{The given line is parallel to the vector}\overrightarrow{b}=3\hat{i}-\hat{j}+2\hat{k}\text{and the given plane is normal to the vector}\overrightarrow{n}=3\hat{i}+4\hat{j}+\hat{k}. \\ \text{We know that the angle θ between the line and the plane is given by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ =\frac{\left(3\hat{i}-\hat{j}+2\hat{k}\right).\left(3\hat{i}+4\hat{j}+\hat{k}\right)}{\left|3\hat{i}-\hat{j}+2\hat{k}\right|\left|3\hat{i}+4\hat{j}+\hat{k}\right|}=\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}=\frac{7}{\sqrt{14}\sqrt{26}}=\frac{7}{\sqrt{2}\sqrt{7}\sqrt{2}\sqrt{13}}=\frac{\sqrt{7}}{\sqrt{52}}=\sqrt{\frac{7}{52}} \\ \Rightarrow \theta ={\sin }^{-1}\left(\sqrt{\frac{7}{52}}\right) \end{gathered}$$