Question

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane 10x + 2y − 11z = 3.

Answer 1

$$\begin{gathered} \text{The given line is parallel to the vector}\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}\text{and the given plane is normal to the vector}\overrightarrow{n}=10\hat{i}+2\hat{j}-11\hat{k}. \\ \text{We know that the angle θ between the line and the plane is given by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ =\frac{\left(2\hat{i}+3\hat{j}+6\hat{k}\right).\left(10\hat{i}+2\hat{j}-11\hat{k}\right)}{\left|2\hat{i}+3\hat{j}+6\hat{k}\right|\left|10\hat{i}+2\hat{j}-11\hat{k}\right|}=\frac{20+6-66}{\sqrt{4+9+36}\sqrt{100+4+121}}=\frac{-40}{\left(7\right)\left(15\right)}=\frac{-8}{21} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{-8}{21}\right) \end{gathered}$$