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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. If the line meets the plane $2 x+y+z=9$ at point $Q$, then the length $P Q$ equals
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Verified Answer
The correct answer is:
$\sqrt{3}$
$\sqrt{3}$
Point $P$ is $(2,-1,2)$
Let this line meet at $Q(h, k, w)$
Direction ratio of this line is
$(h-2, k+1, w-2)$
Since, $d c_s$ are equal $\& d r_s$ are also equal,
So, $h-2=k+1+w-2$
$\Rightarrow \quad k=h-3$ and $w=h$
This line meets the plane
$2 x+y+z=9$ at $Q$, so,
$2 h+k+w=9$ or $2 h+h-3+h=9$
$\Rightarrow 4 h-3=9 \Rightarrow h=3$
and $k=0$ and $w=3$
Distance
$$
\begin{aligned}
P Q & =\sqrt{\left(3-2^2\right)+\left(0-(-1)^2\right)+(3-2)^2} \\
& =\sqrt{1^2+1^2+1^2}=\sqrt{3}
\end{aligned}
$$
Let this line meet at $Q(h, k, w)$
Direction ratio of this line is
$(h-2, k+1, w-2)$
Since, $d c_s$ are equal $\& d r_s$ are also equal,
So, $h-2=k+1+w-2$
$\Rightarrow \quad k=h-3$ and $w=h$
This line meets the plane
$2 x+y+z=9$ at $Q$, so,
$2 h+k+w=9$ or $2 h+h-3+h=9$
$\Rightarrow 4 h-3=9 \Rightarrow h=3$
and $k=0$ and $w=3$
Distance
$$
\begin{aligned}
P Q & =\sqrt{\left(3-2^2\right)+\left(0-(-1)^2\right)+(3-2)^2} \\
& =\sqrt{1^2+1^2+1^2}=\sqrt{3}
\end{aligned}
$$
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