Question

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

Answer 1

We start solving this problem by equating the given equation of the line to some constant $\lambda$. Then we find $x,y$ and $z$ in terms of $\lambda$. Then we substitute these terms in the equation of plane as they intersect, to obtain the value of $\lambda$. Automatically we get the values of $x,y$, and $z$. Hence, we get the coordinates of the point of intersection of the given line and plane. Then we get the angle between the given line and plane by using the formula $\sin \theta =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|}$.

Complete step-by-step solution:
Let us consider the given equation of line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}$.
Now, we equate the above equation to some constant $\lambda$
So, $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}=\lambda$
By the above equation, we get
$\begin{aligned} & x-2=3\lambda \\\ & \Rightarrow x=3\lambda +2..............\left( 1 \right) \\\ & y+1=4\lambda \\\ & \Rightarrow y=4\lambda -1..............\left( 2 \right) \\\ & z-2=2\lambda \\\ & \Rightarrow z=2\lambda +2..............\left( 3 \right) \\\ \end{aligned}$
We were given that the equation of plane is $x-y+z-5=0$.
By substituting the equations (1), (2) and (3) in the above plane equation, we get
$\begin{aligned} & \left( 3\lambda +2 \right)-\left( 4\lambda -1 \right)+\left( 2\lambda +2 \right)-5=0 \\\ & \Rightarrow 3\lambda +2-4\lambda +1+2\lambda +2-5=0 \\\ & \Rightarrow \lambda +5-5=0 \\\ & \Rightarrow \lambda =0 \\\ \end{aligned}$
Now, we substitute the value of $\lambda$ in equation (1), we get
$\begin{aligned} & x=3\lambda +2 \\\ & \Rightarrow x=3\left( 0 \right)+2 \\\ & \Rightarrow x=0+2 \\\ & \Rightarrow x=2 \\\ \end{aligned}$
By substituting the value of $\lambda$ in equation (2), we get
$\begin{aligned} & y+1=4\lambda \\\ & \Rightarrow y+1=4\left( 0 \right) \\\ & \Rightarrow y+1=0 \\\ & \Rightarrow y=-1 \\\ \end{aligned}$
We substitute the value of $\lambda$ in equation (3), we get
$\begin{aligned} & z=2\lambda +2 \\\ & \Rightarrow z=2\left( 0 \right)+2 \\\ & \Rightarrow z=0+2 \\\ & \Rightarrow z=2 \\\ \end{aligned}$
So, we get $x=2,y=-1,z=2$.
Hence, the coordinates of the point of intersection of the given line and plane are $\left( 2,-1,2 \right)$.
Now, let us find the angle between the given line and the plane.
Let us consider the given line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}$.
We can say that the line is parallel to the vector $3\hat{i}+4\hat{j}+2\hat{k}$.
Now, let us consider the given equation of the plane $x-y+z-5=0$.
We can say that the plane is normal to the vector $\hat{i}-\hat{j}+\hat{k}$.
Let us consider the formula for the angle between the line parallel to the vector $\overrightarrow{a}$ and the plane whose normal is $\overrightarrow{b}$, is $\sin \theta =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|}$
By using the above formula, we get
$\sin \theta =\dfrac{\left( 3\hat{i}+4\hat{j}+2\hat{k} \right).\left( \hat{i}-\hat{j}+\hat{k} \right)}{\left| 3\hat{i}+4\hat{j}+2\hat{k} \right|.\left| \hat{i}-\hat{j}+\hat{k} \right|}$
Let us consider the formula, $\left( {{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k} \right).\left( {{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k} \right)={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}$ and $\left| a\hat{i}+b\hat{j}+c\hat{k} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
Using the above formulae, we get
$\begin{aligned} & \sin \theta =\dfrac{\left( 3 \right)\left( 1 \right)+\left( 4 \right)\left( -1 \right)+\left( 2 \right)\left( 1 \right)}{\left( \sqrt{{{3}^{2}}+{{4}^{2}}+{{2}^{2}}} \right)\left( \sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}} \right)} \\\ & \Rightarrow \sin \theta =\dfrac{3-4+2}{\left( \sqrt{29} \right)\left( \sqrt{3} \right)} \\\ & \Rightarrow \sin \theta =\dfrac{1}{\sqrt{87}} \\\ & \Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{87}} \right) \\\ \end{aligned}$
Therefore, the angle between the given line and plane is ${{\sin }^{-1}}\left( \dfrac{1}{\sqrt{87}} \right)$.

Note: One may make a mistake by considering $-2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$ as vector parallel to the line $\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{2}$ instead of taking $3\overrightarrow{i}+4\overrightarrow{j}+2\overrightarrow{k}$. But it is the point on the line not a vector parallel to it, $3\overrightarrow{i}+4\overrightarrow{j}+2\overrightarrow{k}$ is the vector parallel to given line.