Expert: Ahmed Salami - 10/5/2010

Question
Hello I have a question about vectors that seems mathematical yet is in a physics first year course:
The vector -; = Xl + yj + zk, called the position vector,
points from the origin (0. 0, o) to an arbitrary point in space with
coordinaIes (x, y, z). 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 Care constants,lie in a plane
that passes through the origin and that is perpendicular to the vector
Ai + Bj + ck.

I don't know much about 3 dimensional space but I searched and found that for a line to be perpendicular to a plane it has to be perpendicular to two intersecting lines in the plane I also know that for two vectors to be perpendicular there cross product must be 0, but that isn't enough to help me solve this problem .

Is there any way for me to find a vector that lies in the plane Ax+By+Cz=0 ?  I think that will help solve this problem. any ideas?

Answer
Hi Hamad,
First to correct one of your statements. Two vectors are infact orthogonal (i.e perpendicular) if their dot product (and not cross product) is equal to 0.
Now, consider the two vectors;
U = xi + yj + zk
and
V = Ai + Bj + Ck
Their dot product is given by
U.V = Ax + By + Cz
We can thus see that the set of points satisfying the equation Ax + By + Cz = 0 are points lying on the planes containing the vectors U and V such that U.V = 0 i.e perpendicular planes.

Regards