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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.

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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:
The given points are $O(0,0,0)$ and $A(1,2,-3)$
$\Rightarrow\:$ Direction ratio $(d.r.)$ of $OA\: is\:\:(x_2-x_1),(y_2-y_1),(z_2-z_1)$
$i.e.\:\:(1-0),(2-0),(-3-0)=(1,2,-3)$
Since $A$ is foot of perpendicular from origin on the plane,
$OA$ is also $\perp$ to the plane.
$\Rightarrow\:\overrightarrow n=(1,2,-3)$
Step 2:
We know that the equation of 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 $(\overrightarrow n)$=$(1,2,-3)$ and the point $A$ is $(1,2,-3)$.
$\therefore$ The equation of the required plane is $1(x-1)+2(y-2)-3(z+3)=0$
$\Rightarrow\:x-1+2y-4-3z-9=0$
On simplifying
$x+2y-3z-14=0$
Hence equation of required plane is $x+2y-3z-14=0$
answered Feb 3, 2014 by rvidyagovindarajan_1
edited Feb 3, 2014 by rvidyagovindarajan_1
 

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