Question

Find the coordinates of the point where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.

Answer 1

$$\begin{gathered} \text{Let}\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda \text{(say)} \\ \Rightarrow x=3\lambda +2;y=4\lambda -1;z=2\lambda +2...\left(1\right) \\ \text{Since (}\mathit{\text{x}}\text{,}\mathit{\text{y}}\text{,}\mathit{\text{z}}\text{) intersects the plane}x-y+z-5=0, \\ 3\lambda +2-\left(4\lambda -1\right)+2\lambda +2-5=0 \\ \Rightarrow 3\lambda +2-4\lambda +1+2\lambda +2-5=0 \\ \Rightarrow \lambda =0 \\ \text{Substituting this in (1), we get} \\ x=2;y=-1;z=2 \\ \text{So,}\left(x,y,z\right)=\left(2,-1,2\right) \\ \mathbf{\text{Finding the angle}} \\ \text{The given line is parallel to the vector}\overrightarrow{b}=3\hat{i}+4\hat{j}+2\hat{k}\text{and the given plane is normal to the vector}\overrightarrow{n}=\hat{i}-\hat{j}+\hat{k}. \\ \text{We know that the angle θ between a line and a plane is given by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ =\frac{\left(3\hat{i}+4\hat{j}+2\hat{k}\right).\left(\hat{i}-\hat{j}+\hat{k}\right)}{\left|3\hat{i}+4\hat{j}+2\hat{k}\right|\left|\hat{i}-\hat{j}+\hat{k}\right|}=\frac{3-4+2}{\sqrt{9+16+4}\sqrt{1+1+1}}=\frac{1}{\sqrt{87}} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{87}}\right) \end{gathered}$$