# Angle between plane and xy-plane

• May 2nd 2009, 06:50 AM
No Logic Sense
Angle 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 AM
VonNemo19
(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 AM
Plato
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 AM
Soroban
Hello, No Logic Sense!

Quote:

Find the angle between the plane $\displaystyle -900y+165z\:=\:0$ and the $\displaystyle xy$-plane.
Given vectors $\displaystyle \vec u$ and $\displaystyle \vec v$, the angle $\displaystyle \theta$ between them is given by: .$\displaystyle \cos\theta \:=\:\frac{|\vec u \cdot\vec v|}{|\vec v||\vec v|}$

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 AM
No 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 AM
earboth
Quote:

Originally Posted by No 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? :(.....Yes

$\displaystyle \cos(\alpha)=\dfrac{(0,-900,165) \cdot (0,0,1)}{915 \cdot 1} = \dfrac{11}{61}$

which is exactly the same value as posted by Soroban.

But your calculator is switched to the very special mode where a right angle is measured 100°
• May 3rd 2009, 07:31 AM
HallsofIvy
Quote:

Originally Posted by earboth
$\displaystyle \cos(\alpha)=\dfrac{(0,-900,165) \cdot (0,0,1)}{915 \cdot 1} = \dfrac{11}{61}$

which is exactly the same value as posted by Soroban.

But your calculator is switched to the very special mode where a right angle is measured 100°

"grads"- a system used primarily to measure slope of roads has 100 grads per right angle. I once upon a time had a calculator that could be set to "degrees", "radians", or "grads" but I haven't seen one in years.