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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The coordinates of the point on the line through $(3,-4,-5)$ and $(2,-3,1)$ at which the line crosses the plane $2x+y+z=5$ is ?

$\begin{array}{1 1} (a)\:\:(1,-2,7)\:\:\qquad\:\:(b)\:\:(1,-3,8)\:\:\qquad\:\:(c)\:\:(2,-1,4)\:\:\qquad\:\:(d)\:\:(2,-2,5) \end{array} $

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Equation of the line through $P(3,-4,-5) \:\:and\:\:Q(2,-3,1)$ is given by
$\large\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z-1}{-6}=\lambda$
Any point $A$ on this line in terma of $\lambda$ is $A(\lambda+2,-\lambda-3,-6\lambda+1)$
Let this point $A$ be the point where the line crosses the plane $2x+y+z=5$
$\therefore$ the point $A$ satisfies the eqn. of the plane.
$\Rightarrow\:2(\lambda+2)+(-\lambda-3)+(-6\lambda+1)=7$
$\Rightarrow\:\lambda=-1$
$\therefore $ The point $A$ is (1,-2,7)$
answered Jan 2, 2014 by rvidyagovindarajan_1
 

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