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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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True or False: The unit vector normal to the plane $x+2y+3z-6=0$ is $\large \frac{1}{\sqrt {14}}\hat i+\frac{2}{\sqrt {14}}\hat j+\frac{3}{\sqrt {14}}\hat k.$

$\begin{array}{1 1} True \\ False \end{array} $

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Toolbox:
  • Unit vector in the direction of $\overrightarrow a$ is $\hat a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Equation of the given plane is $x+2y+3z-6=0$
The vector equation of the given plane is $\overrightarrow r.(\hat i+2 \hat j+3 \hat k)=6$
The normal vector in $\overrightarrow n$
Unit vector in the direction of $\overrightarrow n$ is $\hat n=\large\frac{\overrightarrow n}{|\overrightarrow n|}$
$|\overrightarrow n|=\sqrt {1^2+2^2+3^2}=\sqrt {14}$
Therefore Unit vector in the direction of $\overrightarrow n$ is
$\large \frac{1}{\sqrt {14}}\hat i+\frac{2}{\sqrt {14}}\hat j+\frac{3}{\sqrt {14}}\hat k.$
Hence the statment is $True$
answered Jun 13, 2013 by meena.p
 

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