Cartesian coordinates in 3D problem.

[Gaz031]

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.

 

[Galileo]

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].

 

[Gaz031]

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

 

[manixc]

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

 

[Gaz031]

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.

 

[Gaz031]

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.

 

[HallsofIvy]

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]

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]

 

[maverick280857]

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.