logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

Find the $\perp$ distance of the point (3,-2,1) from the plane $2x-y+2z+3=0$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Distance of any point $ (x_1,y_1,z_1)$ from the plane $ ax+by+cz+d=0$ is given by $ \bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
Given point $Q(3,-2,1)$ and Eqn. of plane $2x-y+2z+3=0$
We know that the distance of any point $ (x_1,y_1,z_1)$ from the plane $ ax+by+cz+d=0$ is given by
$ \bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
$\perp$ distance =$\bigg|\large\frac{6+2+2+3}{\sqrt {2^2+(-1)^2+2^2}}\bigg|=\frac{13}{3}$
 

 

answered May 22, 2013 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...