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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If O be the origin and the coordinates of P be $(1, 2, – 3)$, then find the equation of the plane passing through P and perpendicular to OP.

$\begin{array}{1 1}(A) x+2y-3z-4=0 \\(B) x+2y+3z-14=0 \\(C) x+2y-3z-14=0 \\ (D) x+2y-3z+14=0 \end{array} $

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Toolbox:
  • Direction ratios of a given vector is whose points are $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $(x_2-x_1),(y_2-y_1),(z_2-z_1)$.
  • Equation of the plane is $a(x-x_1)+b(y_2-y_1)+c(z_2-z_1)=0$
Step 1:
Let the points $o$ be $(0,0,0)$ and $P(1,2,-3)$
Therefore direction ratios of $OP$ are
$(x_2-x_1),(y_2-y_1),(z_2-z_1)$
(i.e)$(1-0),(2-0),(-3-0)$
$\Rightarrow (1,2,-3)$
Step 2:
The plane passing through $(x_1,y_1,z_1)$ is
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Where $a,b,c$ are the direction ratio of normal are $(1,2,-3)$ and the point $P$ is $(1,2,3)$.
Therefore the equation of the required plane is $1(x-1)+2(y-2)+3(z-3)=0$
$x-1+2y-4-3z-9=0$
On simplifying
$x+2y-3z-14=0$
Hence equation of required plane is $x+2y-3z-14=0$
answered Jun 5, 2013 by sreemathi.v
 

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