Question

Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3x − y + z = 1.

Answer 1

$$\begin{gathered} \text{It is given that the line passes through}A\left(3,-4,-2\right)\text{and}B\left(12,2,0\right). \\ \text{So,}\overrightarrow{b}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=12\hat{i}+2\hat{j}+0\hat{k}-\left(3\hat{i}-4\hat{j}-2\hat{k}\right)=9\hat{i}+6\hat{j}+2\hat{k} \\ \text{The given line is parallel to the vector}\overrightarrow{b}=9\hat{i}+6\hat{j}+2\hat{k}\text{and the given plane is normal to the vector}\overrightarrow{n}=3\hat{i}-\hat{j}+\hat{k}. \\ \text{We know that the angle θ between the line and the plane is given by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ =\frac{\left(9\hat{i}+6\hat{j}+2\hat{k}\right).\left(3\hat{i}-\hat{j}+\hat{k}\right)}{\left|9\hat{i}+6\hat{j}+2\hat{k}\right|\left|3\hat{i}-\hat{j}+\hat{k}\right|}=\frac{27-6+2}{\sqrt{81+36+4}\sqrt{9+1+1}}=\frac{23}{11\sqrt{11}} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{23}{11\sqrt{11}}\right) \end{gathered}$$