# Cartesian coordinates in 3D problem.

I've no idea what to do with this, the examples didn't have anything of this style:

The point A has coordinates (3,0,0), the point B has coordinates, (0,3,0), the point C has coordinates (0,0,7). Find, to 0.1 degrees, the sizes of the angle between the planes OAB and ABC, where O is the origin.

Could someone give me an idea of what to do? I've only just introduced myself to cartesian coordinates/vectors in 3 dimensions.

Related Introductory Physics Homework Help News on Phys.org
Galileo
Homework Helper
The plane OAB is simply the xy-plane.
By symmetry you can see (draw a picture) that the angle between the planes is
the same as the angle between the vector [1,1,0] and the vector [-1/2,-1/2,3].

The thing is, i have no idea how to calculate angles in 3d. I always thought of angles as 2 dimensional.

I suggust drawing the points first then sketching the plane OAB and ABC. Then you can treat this as a simple Pythagorean then trig question. As Galileo has stated, the drawing will be in symmetry.

a hint would be finding the length of AB and the line from origin to the midpt of AB

Galileo said:
The plane OAB is simply the xy-plane.
By symmetry you can see (draw a picture) that the angle between the planes is
the same as the angle between the vector [1,1,0] and the vector [-1/2,-1/2,3].
I can't draw 3d shapes but i've sketched it as best as i can. Where did you get the coordinates 1,1,0 and -.5,-.5,3 from? What mathematical process? I hate crappy 3d vectors especially with this crappy book that throws you a question without any method for doing it.

Last edited:
manixc said:
I suggust drawing the points first then sketching the plane OAB and ABC. Then you can treat this as a simple Pythagorean then trig question. As Galileo has stated, the drawing will be in symmetry.

a hint would be finding the length of AB and the line from origin to the midpt of AB
I know the length of AB is root3. The line AB has equation 3y+3x-9=0. Neither do i understand how you can have a three dimensional angle without making a whole lot of extra degrees or using two angles and reinventing trigonometry.

Last edited:
HallsofIvy
Homework Helper
A is on the x-axis and B is on the y-axis. The plane OAB is just the xy-plane.
The plane OAB crosses OAB in the line AB. The "angle" between the planes is the angle between a line in OAB perpendicular to AB and a line in ABC perpendicular to AB, that is between the line through (0,0,7) perpendicular to AB and the line through (0,0,0) perpendicular to AB.

robphy
Homework Helper
Gold Member
Gaz031 said:
The thing is, i have no idea how to calculate angles in 3d. I always thought of angles as 2 dimensional.
Two nonzero, nonparallel vectors (in whatever dimension) determine a plane.
The angle between those vectors is an angle on that plane.

Given nonzero vectors $\vec A$ and $\vec B$, the angle $\theta$ between them can be determined by using two expressions for the dot-product $\vec A \cdot \vec B$.

$$\vec A \cdot \vec B = A_x B_x + A_y B_y +A_z B_z= |\vec A| |\vec B| \cos\theta$$

robphy said:
Two nonzero, nonparallel vectors (in whatever dimension) determine a plane.
The angle between those vectors is an angle on that plane.

Given nonzero vectors $\vec A$ and $\vec B$, the angle $\theta$ between them can be determined by using two expressions for the dot-product $\vec A \cdot \vec B$.

$$\vec A \cdot \vec B = A_x B_x + A_y B_y +A_z B_z= |\vec A| |\vec B| \cos\theta$$
Sometimes the angle between two planes is determined by first computing the angle between the normal vectors to a plane. The normal to a plane $$\vec n$$ is simply a vector perpendicular to the plane. The vector equation of a plane (fyi) is written concisely as

$$\vec n \cdot \vec r = 0$$

Here $$\vec r$$ is a vector perpendicular to n, which means r lies in the plane.