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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane passing through the point $( -1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$.

$\begin{array}{1 1} (A) x-3y-2z+25=0 \\(B) x-3y-2z-25=0 \\ (C) 7x-8y+3z-25=0 \\ (D) 7x-8y+3z+25=0 \end{array} $

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Toolbox:
  • Equation of a plane,whose direction ratios are $(a,b,c)$ and points $(x_1,y_1,z_1)$ is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
  • If two planes of $\perp$ then
  • $a_1a_2+b_1b_2+c_1c_2=0$
Step 1:
Let the equation of the plane passing through the point $(-1,3,2)$ be
$a(x+1)+b(y-3)+c(z-2)=0$------(1)
Here $a,b,c$ are the direction ratios of normal to the plane.
If two planes are $\perp$ then
$a_1a_2+b_1b_2+c_1c_2=0$
Plane (1) is $\perp$ to the plane $x+2y+3z=5$
The direction cosines are $(1,2,3)$
Therefore $a.1+b.2+c.3=0$
$\Rightarrow a+2b+3c=0$------(2)
Step 2:
Also it is given plane (1) is $\perp$ to $3x+3y+z=0$
The direction cosines are $(3,3,1)$
Therefore $a.3+b.3+c.1=0$
$\Rightarrow 3a+3b+c=0$------(3)
Step 3:
On solving equ(2) and equ(3) we get
$\large\frac{a}{2\times 1-3\times 3}=\frac{b}{3\times 3-1\times 1}=\frac{c}{1\times 3-2\times 3}$
(i.e) $\large\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}$
Step 4:
Let this be equal to $k$
$\large\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=$$k$
$a=-7k$
$b=8k$
$c=-3k$
Step 5:
Substituting the value of $a,b$ and $c$ in equ(1) we get
$-7k(x+1)+8k(y-3)-3k(z-2)=0$
$-7kx-7k+8ky-24k-3kz+6k=0$
$k[-7x-7+8y-24-3z+6]=0$
$\Rightarrow -7x-7+8y-24-3z+6=0$
$-7x+8y-3z-25=0$
$7x-8y+3z+25=0$
Hence this is the required equation of the plane.
answered Jun 4, 2013 by sreemathi.v
edited Jun 4, 2013 by sreemathi.v
 

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