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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the vector and cartesian equations of the planes that passes through the point $(1, 0, -2)$ and the normal to the plane is $ \hat i+\hat j - \hat k $

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

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Toolbox:
  • Vector equation of a plane passing through a point and normal to the plane is given by $(\overrightarrow r-\overrightarrow a).\overrightarrow N=0$
  • Cartesian equation is given by $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
Step 1:
The position vector of point $(1,0,-2)$ is $\overrightarrow a=\hat i-2\hat k$
The normal vector $\overrightarrow N$ perpendicular to the plane is $\overrightarrow N=\hat i+\hat j-\hat k$
We know the vector equation of the plane is given by $(\overrightarrow r-\overrightarrow a).\overrightarrow N=0$
Step 2:
On substituting for $\overrightarrow a$ and $\overrightarrow N$ we get,
$[\overrightarrow r-(\hat i-2\hat k)].(\hat i+\hat j-\hat k)=0$
But we know $\overrightarrow r=x\hat i+y\hat j+z\hat k$
Therefore $[(x\hat i+y\hat j+z\hat k)-(\hat i-2\hat k)].(\hat i+\hat j-\hat k)=0$
Step 3:
On simplifying we get,
$[(x-1)\hat i+y\hat j+(z+2)\hat k)].(\hat i+\hat j-\hat k)=0$
We know that $\hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1$
$\Rightarrow (x-1)+y-(z+2)=0$
$x+y-z-3=0$
$x+y-z=3$
Therefore $x+y-z=3$ is the required Cartesian equation of the plane.
answered May 31, 2013 by sreemathi.v
 

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