I have a plane with the equation -900y+165z=0 and I want to find the angle between this plane and the xy-plane. What do I do?
I'm not sure that I understand your question. It looks like all you have is the equation to a line in the zy plane, which is, by definition, perpindicular to the xy plane. Now, we could perhaps find the angle of the line that goes through the zy plane by anylizing its slope which is 900/165 (I don't feel like reducing). Does this help you?
Hello, No Logic Sense!
Given vectors $\displaystyle \vec u$ and $\displaystyle \vec v$, the angle $\displaystyle \theta$ between them is given by: .$\displaystyle \cos\theta \:=\:\frac{|\vec u \cdot\vec v|}{|\vec v||\vec v|} $Find the angle between the plane $\displaystyle -900y+165z\:=\:0$ and the $\displaystyle xy$-plane.
The angle between their normal vectors is the angle between the planes.
The first plane has normal vector: .$\displaystyle \vec u \:=\:\langle 0,\text{-}900,165\rangle \:=\:\langle0,\text{-}60,11\rangle $
. . Its magnitude is: .$\displaystyle |\vec u| \:=\:\sqrt{0^2 + (\text{-}60)^2+9^2} \:=\:\sqrt{3721} \:=\:61$
The $\displaystyle xy$-plane has normal vector: .$\displaystyle \vec v \:=\:\langle0,0,1\rangle$
. . It is a unit vector: .$\displaystyle |\vec v| \:=\:1$
Hence: .$\displaystyle \cos\theta \;=\;\frac{|\langle0,\text{-}60,11\rangle\cdot\langle0,0,1\rangle|}{61\cdot1} \;=\;\frac{0+0+11}{61} \;=\;\frac{11}{61}$
Therefore: .$\displaystyle \theta \;=\;\arccos\left(\tfrac{11}{61}\right) \;\approx\;79.6^o $