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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane$x -y + z = 0.$

$\begin{array}{1 1} x+z+2=0 \\ x-z-2=0 \\ x-z+2=0 \\ x-y+2=0\end{array} $

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1 Answer

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Toolbox:
  • Equation of the plane passing through the intersection of the plane is given by
  • $(a_1x+b_1y+c_1z+d_1)+\lambda(a_2x+b_2y+c_2z+d_2)=0$
Step 1:
Given equation of the planes are:
$x+y+z=1$ and $2x+3y+4z=5$
Hence the equation of the plane through the intersection of the planes is
$(x+y+z-1)+\lambda(2x+3y+4z-5)=0$-------(1)
On simplifying we get
$(2\lambda+1)x+(3\lambda+1)y+(4\lambda+1)z-(5\lambda+1)=0$
Step 2:
Let the direction ratios be $a_1,b_1$ and $c_1$ which are $(2\lambda+1),(3\lambda+1)$ and $(4\lambda+1)$
It is given that the plane is $\perp$ to $x-y+z=0$
The direction ratios of this plane be $a_2,b_2$ and $c_2$ which are $(1,-1,1)$.
Since the planes are perpendicular
$a_1a_2+b_1b_2+c_1c_2=0$
$(i.e)1.(2\lambda+1)+(-1)(3\lambda+1)+4\lambda+1)=0$
Step 3:
On simplifying we get,
$2\lambda+1-3\lambda+1+4\lambda+1=0$
$\Rightarrow 3\lambda+1=0$
Therefore $\lambda=-\large\frac{1}{3}$
Substituting this value of $\lambda$ in equ(1) we get
$(x+y+z-1)+(\large\frac{-1}{3})($$2x+3y+4z-5)=0$
Step 4:
On simplifying we get,
$\large\frac{1}{3}$$x-\large\frac{1}{3}$$z+\large\frac{2}{3}$$=0$
$\Rightarrow x-z+2=0$
This is required equation of the plane.
answered Jun 3, 2013 by sreemathi.v
 

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