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?

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- May 2nd 2009, 06:50 AMNo Logic SenseAngle between plane and xy-plane
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?

- May 2nd 2009, 07:30 AMVonNemo19
(Wondering)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?

- May 2nd 2009, 07:53 AMPlato
The normal to the given plane is $\displaystyle <0,-900,165>$ and the normal to the xy-plane is $\displaystyle <0,0,1>$.

What is the angle between those two vectors? - May 2nd 2009, 08:01 AMSoroban
Hello, No Logic Sense!

Quote:

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 $

- May 2nd 2009, 09:45 AMNo Logic Sense
Ahaaa, but my normal vector for the plane is:

(0;-900;165)

and then the xy-plane has the (0;0;1)

I get the angle to 88,24 degrees, is that wrong? :( - May 2nd 2009, 11:11 AMearboth
- May 3rd 2009, 07:31 AMHallsofIvy