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

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat i + 5\hat j - 6 \hat k.\)

1 Answer

Toolbox:
  • $\hat n=\large\frac{\overrightarrow n}{|\overrightarrow n|}$
  • The equation of the plane with position vector $\overrightarrow r$ is given by $\overrightarrow r.\hat n=d$
Step 1:
The normal vector given is $\overrightarrow n=3\hat i+5\hat j-6\hat k$
The unit vector is given by $\hat n=\large\frac{\overrightarrow n}{|\overrightarrow n|}$
$\hat n=\large\frac{3\hat i+5\hat j-6\hat k}{\sqrt{3^2+5^2+(-6)^2}}$
$\quad=\large\frac{3\hat i+5\hat j-6\hat k}{\sqrt{9+25+36}}$
$\quad=\large\frac{3\hat i+5\hat j-6\hat k}{\sqrt{70}}$
Step 2:
We know that the equation of the plane with position vector $\overrightarrow r$ is given by $\overrightarrow r.\hat n=d$
$\Rightarrow \overrightarrow r.\big(\large\frac{3\hat i+5\hat j-6\hat k}{\sqrt {70}}\big)$$=7$
This is the required vector equation of the required plane.
answered May 30, 2013 by sreemathi.v
 

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