JEE Advanced · Mathematics · 31. 3D Geometry
A line with positive direction cosines passes through the point \(P(2,-1,2)\) and makes equal angles with the coordinate axes. The line meets the plane \(2 x+y+z=9\) at point \(Q\). The length of the line segment \(P Q\) equals
- A
1
- B
\(\sqrt{2}\)
- C
\(\sqrt{3}\)
- D
2
Answer & Solution
Correct Answer
(C)
\(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\[
\text { Since, } l=m=n=\frac{1}{\sqrt{3}}
\]

\(\therefore\) Equation of line are
\[
\begin{gathered}
\frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\
\Rightarrow \quad x-2=y+1=z-2=r \text { (say) }
\end{gathered}
\]
\(\therefore\) Any point on the line is
\[
Q \equiv(r+2 r-1, r+2)
\]
\(\because Q\) lies on the plane \(2 x+y+z=9\)
\[
\begin{aligned}
& \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\
& \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\
& \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\
& =\sqrt{3}
\end{aligned}
\]
\text { Since, } l=m=n=\frac{1}{\sqrt{3}}
\]

\(\therefore\) Equation of line are
\[
\begin{gathered}
\frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\
\Rightarrow \quad x-2=y+1=z-2=r \text { (say) }
\end{gathered}
\]
\(\therefore\) Any point on the line is
\[
Q \equiv(r+2 r-1, r+2)
\]
\(\because Q\) lies on the plane \(2 x+y+z=9\)
\[
\begin{aligned}
& \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\
& \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\
& \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\
& =\sqrt{3}
\end{aligned}
\]
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