I suggest that you read post 17.2 before this.
In post 17.2 we saw how to add two-dimensional vectors – vectors that could be drawn on a sheet of paper. But forces can act in three dimensions – up-and-down, as well as north-south and east-west. A three-dimensional force has the same relationship to a two-dimensional force as a sculpture has to a picture.
How do we do vector algebra (see post 17.2) in three-dimensions? We start off by defining a three-dimensional orthogonal Cartesian coordinate system; we met a system like this, but in two dimensions, in appendix 2 of post 16.50. Now we have an X-axis and a Y-axis, that are perpendicular to each other and cross at an origin. But we add a third axis, the Z-axis, that is perpendicular to both the X-axis and the Y-axis. When we add the Z-axis, we need to be careful because three-dimensional Cartesian coordinate systems can be either right-handed or left-handed.
We usually construct right-handed coordinate systems. To do this, curl the fingers of your right hand – so your fingers point from the X-axis to the Y-axis. If you point your thumb up, it then points in the Z-axis direction; in the picture on the right-hand side, the Z-axis points towards you, out of the plane of the paper.
In the same way as we defined unit vectors i and j in the X and Y-axes directions (post 17.2), we now define a unit vector k in the Z-axis direction. Remember that a unit vector simply defines a direction and has a length equal to the number one and no units (like metres or newtons).
So we can write the vector representing a three-dimensional force as
F = iFx + jFy + kFz.
Here Fx, Fy and Fz are the components of F in the X, Y and Z-axis directions. The magnitude of F is given, by Pythagoras’ theorem (see the appendix at the end of this post for more explanation), by
F2 = Fx2 + Fy2 + Fz2 so that F = √(Fx2 + Fy2 + Fz2).
There are two related ways in which we can use angles to specify the direction of F. We will use a simple extension of the method we used for two-dimensional vectors by defining the angle φ that F makes with the Z-axis. Now we define θ as the angle between the projection of F on to the XY-plane and the X-axis (see picture above). From the definitions of cosine and sine (appendix 3 of post 16.50)
Fx = F sin φ cos θ, Fy = F sin φ sin θ and Fz = F cos φ.
We can calculate the angles θ and φ from the components of the vectors. The angle θ is calculated in exactly the same way as for a two-dimensional vector (equation 5 of post 17.2). The angle φ is calculated from
φ = arccos(Fz/F)
where arccos(x) just means “the angle whose cosine is x”. You can look up arccos values at: http://www.rapidtables.com/calc/math/Arccos_Calculator.htm.
The resultant (sum) of two vectors
F = iFx + jFy + kFz and f = ifx + jfy + kfz.
is then given by
R = F + f = i(Fx + fx) + j(Fy + fy) + k(Fz + fz).
So, the algebra of three-dimensional vectors is a simple extension of the algebra of two-dimensional vectors. We can even extend the idea of a vector into 4 or more dimensions; but it is then easier to represent vectors using the notation of matrix algebra which is a bit more difficult to explain.
The purpose of this appendix is to show that F2 = Fx2 + Fy2 + Fz2.
In the figure, the brown line is the projection (see appendix 3 of post 16.50) of the vector F on to the XY-plane.
According to Pythagoras’ theorem, the length of this line is √(Fx2 + Fy2). Here Fx is the length of the blue line and Fy is the length of the red line. The statements in this paragraph are explained and proved in post 17.2.
The green line is the projection of F on to the Z-axis and is denoted by Fz; the dashed green line is parallel to it and has the same length.
F, the dashed green line and the brown line form a right-angled triangle. So, by Pythagoras’ theorem
F2 = (√(Fx2 + Fy2))2 + Fz2 = Fx2 + Fy2 + Fz2.