Question

A line with positive direction cosines passes through the point $P\left( 2,-1,2 \right)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals
A. $1$
B. $\sqrt{2}$
C. $\sqrt{3}$
D. $2$

Answer 1

Hint: In this problem we need to calculate the length of the given segment according to the given geometric conditions. In the question we have given that the line has positive directional cosines and makes equal angles with the coordinate system. So, the directional cosines of the line will become $l=m=n=\dfrac{1}{\sqrt{3}}$. Now we have the point $P\left( 2,-1,2 \right)$ and the directional cosines, so we can write the equation of the line as $\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n}=r$. From this we can write the equation of the line and calculate the coordinates of the point $Q$ which lies on the calculated line. Now in the problem they have mentioned that the point $Q$ also lies on the plane $2x+y+z=9$. So, we will substitute the coordinates of the point $Q$ and calculate the value of $r$. From this we can simply find the distance between two points by using the formula$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$.

Complete step by step answer:
Given that, A line with positive direction cosines passes through the point $P\left( 2,-1,2 \right)$.
If a line is making equal angle with the all the coordinate axis, then the directional cosines of the line will be $\left( l,m,n \right)=\left( \pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}} \right)$. But in the problem, they have mentioned that the given line has positive directional cosines, then the dc’s of the line will be
$l=m=n=\dfrac{1}{\sqrt{3}}$.
Now the equation of the line having the directional cosines as $l=m=n=\dfrac{1}{\sqrt{3}}$ and passes through the point $P\left( 2,-1,2 \right)$ from the know formula $\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n}=r$ is
$\begin{align}
  & \dfrac{x-2}{\dfrac{1}{\sqrt{3}}}=\dfrac{y+1}{\dfrac{1}{\sqrt{3}}}=\dfrac{z-2}{\dfrac{1}{\sqrt{3}}}=r \\
 & \Rightarrow x-2=y+1=z-2=r \\
\end{align}$
Where $r$ is a constant.
Now the coordinates of the point $Q$ which lies in the above line is given by
$Q=\left( r+2,r-1,r+2 \right)$
But the point $Q$ also lies in the plane $2x+y+z=9$, then we can write
$\begin{align}
  & 2\left( r+2 \right)+\left( r-1 \right)+\left( r+2 \right)=9 \\
 & \Rightarrow 2r+4+r-1+r+2=9 \\
 & \Rightarrow 4r+5=9 \\
 & \Rightarrow 4r=4 \\
 & \Rightarrow r=1 \\
\end{align}$
From the value of $r$ the coordinates of the point $Q$ will be
$\begin{align}
  & Q=\left( 1+2,1-1,1+2 \right) \\
 & \Rightarrow Q=\left( 3,0,3 \right) \\
\end{align}$
Now the distance between the two points $P\left( 2,-1,2 \right)$ and $Q\left( 3,0,3 \right)$ from the formula $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$ is given by
$\begin{align}
  & d=\sqrt{{{\left( 3-2 \right)}^{2}}+{{\left( 0+1 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{3} \\
\end{align}$

Hence the distance between the points $P\left( 2,-1,2 \right)$ and $Q\left( 3,0,3 \right)$ is $\sqrt{3}$ units.

Note: We can also simplify solve this problem by assuming the coordinates of the point $Q$ as $\left( 2+\dfrac{d}{\sqrt{3}},-1+\dfrac{d}{\sqrt{3}},2+\dfrac{d}{\sqrt{3}} \right)$ where $d$ is the distance between the two points. Now on substituting the point $Q$ in the plane $2x+y+z=9$ we will get
$\begin{align}
  & 2\left( 2+\dfrac{d}{\sqrt{3}} \right)+\left( -1+\dfrac{d}{\sqrt{3}} \right)+\left( 2+\dfrac{d}{\sqrt{3}} \right)=9 \\
 & \Rightarrow 4+\dfrac{2d}{\sqrt{3}}-1+\dfrac{d}{\sqrt{3}}+2+\dfrac{d}{\sqrt{3}}=9 \\
 & \Rightarrow \dfrac{4d}{\sqrt{3}}=4 \\
 & \Rightarrow d=\sqrt{3} \\
\end{align}$
From both the methods we got the same result.