How Do You Prove a Vector is Perpendicular to a Plane?

AI Thread Summary
To prove a vector is perpendicular to a plane, it's essential to understand that the equation Ax + By + Cz = 0 defines a plane through the origin, with the normal vector n = Ai + Bj + Ck. The position vector r = xi + yj + zk points to any point (x, y, z) in space. By demonstrating that the dot product r · n equals zero, it confirms that the vector r is perpendicular to the normal vector n, indicating that r lies in the defined plane. The relationship between the vector and the plane is established through the equation r · n = 0, confirming the perpendicularity. This understanding clarifies both the nature of the plane and the conditions for perpendicularity.
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Homework Statement


The vector r=xi+yj+zk, called the position vector, points from the origin(0,0,0) to an arbitrary point in space with coordinates(x,y,z).


Homework Equations


Use what you know about vectors to prove the following: All points (x,y,z) that satisfy the equation Ax+By+Cz=0 where A,B and C are constants,lie in a plane that passes through the origin and that is, perpendicular to the vector n=Ai+Bj+Ck.


The Attempt at a Solution


I can only show that the position vector pointing at (x,y,z) is perpendicular to n=Ai+Bj+Ck.. I did this by using cosa= (Ax+By+Cz)/rn to find the angle between n and r. Ax+By+Cz=0 so cosa=0 and a=90 degrees. But i can't still show that n is perpendicular to the plane. I'm also confused if the problem also asks to show that Ax+By+Cz=0 is a plane.Can somebody please help me. Thanks
 
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Well, the plane is defined as \vec{r} \cdot \vec{n} = 0. This states that r is perpendicular to n. So every point r for which this equation holds, lies in the plane. See also this page on planes: http://mathworld.wolfram.com/Plane.html. Hope this helps!
 
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