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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of the plane passing through the line of intersection of the planes $ 2x+y-z=3 \: and \: 5x-3y+4z+9=0$ and parallel to the line \(\large \frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}. \)

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Toolbox:
  • Vector equation of a line passing through a point and parallel to a given vector is $\overrightarrow r=\overrightarrow a+\lambda \overrightarrow b$
Step 1:
Equation of the planes are $2x+y-z=3$ and $5x-3y+4z+9=0$
The equation of the plane passing through the line of intersection of these planes is
$(2x+y-z-3)+\lambda(5x-3y+4z+9)=0$
$x(2+5\lambda)+y(1-3\lambda)+2(4\lambda-1)+9\lambda-3=0$-----(1)
Step 2:
The plane is parallel to the line
$\large\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}$
$\therefore 2(2+5\lambda)+4(1-3\lambda)+5(4\lambda-1)=0$
(i.e) $18\lambda+3=0$
$\Rightarrow \lambda=-\large\frac{1}{6}$
Step 3:
Put the value of $\lambda$ in emu(1) we obtain
$x(2-\large\frac{5}{6})+y(1+\large\frac{3}{6})+z(-\large\frac{4}{6}-1)-\large\frac{9}{6}-3=0$
$\Rightarrow 7x+9y-10z-27=0$
This is the equation of the required plane.
answered Nov 12, 2013 by sreemathi.v
 

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