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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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If the plane $x-2y+3z=0$ is rotated through a right angle about its line of intersection with the plane $2x+3y-4z-5=0$, then the eqn. of the plane in its new position is ?

$\begin {array}{cc} (a)\:28x-17y+9z=0\: & \:(b)\:22x+5y-4z-35=0 \\ \:(c)\:25x+17y+52z-25=0\:& (d)\:x+35y-10z-70=0\end {array} $

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1 Answer

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Given equations of the given plane are $x-2y+3z=0$..........(i) and
$2x+3y-4z-5=0$.........(ii)
Equation of the plane through the line of intersection of (i) and (ii) is given by
$x-2y+3z+\lambda(2x+3y-4z-5)=0$.
$=x(1+2\lambda)+(y(-2+3\lambda)+z(3-4\lambda)-5\lambda=0$...........(iii)
The equation of the $x-2y+3z=0$ in its new position given by (iii).
But the new position is got by rotating (i) by $90^{\circ}$
$\therefore$ The angle between (i) and (iii) is $90^{\circ}$
$\Rightarrow\:(1,-2,3).(1+2\lambda,\:\:-2+3\lambda,\:\:3-4\lambda)=0$
$\Rightarrow\:1+2\lambda+4-6\lambda+9-12\lambda=0$
$\Rightarrow\:\lambda=\large\frac{7}{8}$
$\therefore$ By substituting $\lambda$ in the required eqn. of the plane (iii) it becomes
$22x+5y-4z+35=0$
answered Jan 10, 2014 by rvidyagovindarajan_1
 

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