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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane which is perpendicular to the plane $5x+3y+6z+8=0$ and which contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0.$

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Toolbox:
  • Equation of a plane which passes through the line intersection of two planes is $(a_1x+b_1y+c_1z+d_1)+\lambda (a_2 x+b_2 y+c_2 z+d_2)$
  • If two planes are perpendicular then $a_1a_2+b_1b_2+c_1c_2=0$ where $(a_1,b_1,c_1)$ and $ (a_2,b_2,c_2)$ are the direction of the two planes respectively.
Step 1:
Let the equation of a plane passing through the line of intersection of the planes $(x+2y+3z-4)=0$ and $(2x+y-z+5)=0$ be
$(x+2y+3z-4)+\lambda (2x+y-z+5)=0$
=> $ x(1+2 \lambda)+y(2 +\lambda)+z(3 -\lambda)-4+5 \lambda=0$ -----(1)
This is perpendicular to the plane $5x+3y+6z+8=0$
$(a_1a_2+b_1b_2+c_1c_2)=0$
Therefore $5(1+2 \lambda)+3 (2+\lambda)+6(3-\lambda)=0$
On simplifying we get,
$5+10 \lambda+6+3 \lambda+18-6 \lambda=0$
$=>7 \lambda+29=0$
Therefore $ \lambda=\large\frac{-29}{7}$
Step 2:
Substituting for $\lambda$ in equ (1) we get,
$x\bigg[1+2 .\bigg(\large\frac{-29}{7}$$\bigg)\bigg]+y\bigg[2 +\bigg(\large\frac{-29}{7}$$\bigg)\bigg]+z\bigg[3- \bigg(\large\frac{-29}{7}$$\bigg)\bigg]-4+5\bigg(\large\frac{-29}{7}\bigg)=0$
On simplifying we get,
$\large\frac{-51}{7}x-\frac{15}{7} y+\frac{50}{7} z-\frac{173}{7}=0$
$=>51 x+15 y-50z+173=0$
This is the required solution of the plane.
answered Jun 12, 2013 by meena.p
 

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