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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry

Find the Cartesian equation of the plane $(b)\; \overrightarrow r . (2\hat i +3\hat j - 4\hat k) = 1 $

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  • For any arbitrary point $P(x,y,z)$ on the plane,the position vector $\overrightarrow r$ is given by $\overrightarrow r=x\hat i+y\hat j+z\hat k$
  • The Cartesian equation of the plane is of the form $lx+my+nz=d$
Step 1:
The given equation of the plane is $\overrightarrow r.(2\hat i+3\hat j-4\hat k)=1$-----(1)
We know that $\overrightarrow r=x\hat i+y\hat j+z\hat k$
Substituting for $\overrightarrow r$ in equ(1) we get,
$(x\hat i+y\hat j+z\hat k).(2\hat i+3\hat j-4\hat k)=1$
Step 2:
We know that $\hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1$
$\Rightarrow 2x+3y-4z=1$
This equation is the required Cartesian equation of the plane.
answered May 30, 2013 by sreemathi.v

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