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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
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Verified Answer
The correct answer is:
$\sqrt{3}$
$\sqrt{3}$
$$
\text { Since, } l=m=n=\frac{1}{\sqrt{3}}
$$
$\therefore$ 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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