3D Geometry – How to Find the Angle of a Surface or 3D Plane

3danglegeometry

I have a plane with equation 3x + 2y + 5z – 19 = 0. With the help of Python/MatPlotlib, I could draw this plane also. Please find the image below.

What I would like to know is how to determine the angle of this plane?

Best Answer

The normal vector to the plane is $n=(3,2,5)$, from that we can find the angles we are interested in using dot product.

For example the angle for the normal with $x-y$ plane, with normal $e_3=(0,0,1)$ is given by

$$\cos \theta=\frac{n\cdot e_3}{|n||e_3|}=\frac{5}{\sqrt{28}}$$

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[Math] Extend angle between two 3D vectors to x-y plane.

If I read your diagram correctly, the plane defined by vectors $v_1$ and $v_2$ intersects the $x$-$y$ plane at a vector that I'll call $v_3$, the magnitude of which is irrelevant. You want the angle from $v_3$ to $v_2$ and also the angle from $v_3$ to $v_1$. So you can define $v_3 = v_1 + \alpha v_2$ where $\alpha$ is such that the $z$-component of $v_3$ is zero. This value $\alpha$ is well defined given that the $z$-component of $v_2$ is not zero. Once you find $v_3$, you know that you will use the dot product to find the angles.

(At first I thought, if you wanted the angle between the two vectors' projections onto the XY-plane, you could just set each z-component to zero and take the dot product. Of course there is no such angle if either vector has x-component and y-component equal to zero.)

[Math] Z Intersection of Cylinder and Plane Base On Angle Around Cylinder

For any point $(x, y, z)$ on the green plane, you have $$ \tan \theta = \frac{z}{x}. $$ hence $z = x \tan(\theta)$.

Letting $t = \tan \theta$, to save a little typing, this means that your third line of code should be something like

function z_cylinder_plane_intersection(y_angle, z_angle, radius) = 
            (radius * cos(z_angle)) * tan(theta)

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Related Solutions

[Math] Extend angle between two 3D vectors to x-y plane.

[Math] Z Intersection of Cylinder and Plane Base On Angle Around Cylinder

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