• Support PF! Buy your school textbooks, materials and every day products Here!

Cartesian coordinates in 3D problem.

  • Thread starter Gaz031
  • Start date
  • #1
51
0
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.
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
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].
 
  • #3
51
0
The thing is, i have no idea how to calculate angles in 3d. I always thought of angles as 2 dimensional.
 
  • #4
3
0
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
 
  • #5
51
0
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:
  • #6
51
0
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:
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,770
911
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.
 
  • #8
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,462
680
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 [itex]\vec A[/itex] and [itex]\vec B[/itex], the angle [itex]\theta[/itex] between them can be determined by using two expressions for the dot-product [itex]\vec A \cdot \vec B[/itex].

[tex]\vec A \cdot \vec B = A_x B_x + A_y B_y +A_z B_z= |\vec A| |\vec B| \cos\theta [/tex]
 
  • #9
1,789
4
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 [itex]\vec A[/itex] and [itex]\vec B[/itex], the angle [itex]\theta[/itex] between them can be determined by using two expressions for the dot-product [itex]\vec A \cdot \vec B[/itex].

[tex]\vec A \cdot \vec B = A_x B_x + A_y B_y +A_z B_z= |\vec A| |\vec B| \cos\theta [/tex]
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 [tex]\vec n[/tex] is simply a vector perpendicular to the plane. The vector equation of a plane (fyi) is written concisely as

[tex]\vec n \cdot \vec r = 0[/tex]

Here [tex]\vec r[/tex] is a vector perpendicular to n, which means r lies in the plane.
 

Related Threads for: Cartesian coordinates in 3D problem.

Replies
1
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
4
Views
329
Replies
1
Views
1K
Replies
1
Views
8K
  • Last Post
Replies
12
Views
789
Top