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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  3-D Geometry

A line with positive direction cosines passes through the point $ P(2,-1,2)$ and makes equal angle with coordinate axes. The line meets the plane $2x+y+z=9$ at the point $Q$. The length of the segment $PQ$=?

$\begin{array}{1 1} 1 \\ \sqrt 2 \\ \sqrt 3 \\ 2 \end{array} $

1 Answer

Given the line makes equal angle with the coordinate axes .
$\therefore\:d.c. $ of the line = $(\large\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})$
$\Rightarrow\:d.r$ of the line is $(1,1,1)$
The line passes through $P(2,-1,2)$
$\therefore$ The eqn. of the line is $\large\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}$$=\lambda$
$\Rightarrow\:$ Any point on the line say $Q$ is $(\lambda+2,\lambda-1,\lambda+2)$
Let this point be the point where the line cuts the plane $2x+y+z-9=0$
$\Rightarrow\:Q$ satisfies the eqn. of the plane.
$\Rightarrow\:2(\lambda+2)+(\lambda-1)+(\lambda+2)-9=0$
$\Rightarrow\:4\lambda-4=0\:\:or\:\:\lambda=1$
$\therefore\:Q(3,0,3)$ and $PQ=\sqrt {1+1+1}=\sqrt 3$
answered Dec 23, 2013 by rvidyagovindarajan_1
 

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