3D Geometry - Why do we require all three Direction Angles?

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I have already talked about direction angles in great detail in my previous post. If you are here to read about the essentials I’ll suggest you should read my previous post. 

In this short post I want to show you a neat little visualisation, something which, despite my best efforts, I couldn't squeeze into earlier post. I’ll intend to keep it short and to the point as possible, because I have already written a great deal about direction angles. So let’s get straight to it.

Firstly, let me state the definition(sort of). Direction angles of a vector are the three angles made by the vector with positive arms of X, Y and Z axes. 

Direction Angles of   :   60°, 45°, 60°

          \(\vec{u}\) 

Direction angles are a way to express the direction of a vector in 3D. We can think of them as the three coordinates for the direction of the vector. Just like we need all three X, Y and Z coordinates of a point to describe its location in 3D, we need all three direction angles to properly describe the direction of the vector. Only one or only two direction angles won’t be enough. 

Let’s say there is a point named P in some three dimensionsional space. We don’t know its location yet, it could be anywhere in space. But if we are given that its X-coordinate is 3, we know that P will be any one of infinitely many points on the plane X = 3. Because every point on the plane X = 3 has X-coordinate 3.

If we are also given that its Y-coordinate is 4, then it further reduces the set of all possible locations of P down to the set of all points that lie on the line parallel to Z-axis, 

\(X\hat{i} + Y\hat{j} + Z\hat{k}\) = \((3\hat{i} + 4\hat{j} + \hat{k})\) + \(\lambda\hat{k}\) \(\lambda\) ∈ R, 

on the plane. Every point on this line has 2 and 4 as its X and Y coordinates.  

If we know the complete set of all three coordinates of P (3,2,3), we can easily pinpoint its exact location in space. 

In a similar fashion, we can also explain the need for all three direction angles to describe the direction of a vector from a nice little piece of geometry. 

Let’s say there is a \(\vec{u}\) in 3D of magnitude 4 units. Geometrically, \(\vec{v}\) is an arrow of length 4 units. To begin with, the vector could be directed anywhere in 3D. We don’t know its direction yet. 

Suppose we know that its direction angle that it makes with the positive X-axis is 60°. 

Think about all possible arrows of size 4 units from the origin that are making a 60° angle with positive X-axis. It is a good brain exercise. 

In fact, you might be surprised to hear that all possible vector arrows will together form a familiar 3D shape - the hollow right circular cone. This is similar to how the set of all points with X-coordinate as 3 forge the plane X = 3.  

Imagine some directed line segment making a 60° angle with a fixed line. See in the figure below. 

Keeping its one end fixed, rotate the other end of the segment around the fixed line in one full circle, the resulting 3D shape will be the hollow right circular cone.

Focus on the parameters that are changing during the rotation. The length of the directed line segment will always remains fixed throughout rotation, so will the 60° angle between the segment and the fixed line. Only thing worth noting that is constantly changing during rotation is the direction of the segment. 

If the directed line segment was our \(\vec{u}\) of magnitude 4 units, making 60° angle with the positive arm of the X-axis(fixed line), then \(\vec{u}\) could be any one of the infinite arrows, all diverging away from the positive X-axis along the lateral surface of the cone as shown in the figure below. 

Or we could also say that the span of all possible arrows of length 4 units from the origin that form a 60° angle with positive X-axis, is a right circular cone. 

If we were given the direction angle of 45° with the positive arm of the Y-axis instead, then again, the span of all possible arrows of length 4 units from the origin that form a 45° angle with positive Y-axis is the right circular cone as shown below. 

Ok. So one direction angle is clearly not sufficient. But what if we are given both these direction angles? 

Well in that case, the vector has to be on the lateral surfaces of both the cones above. The two cones are intersecting in two line segments. So the arrow representing \(\vec{u}\) could be any one of the two arrows from the origin along the two common line segments as shown below. 

We see the information of two direction angles reduces the set of infinite possible directions of \(\vec{u}\) down to just two possible cases. But we still need the third direction angle with the positive Z-axis to tell which of the two arrows is \(\vec{u}\). 

The arrow that is above the XY-plane in the figure above, is making an acute angle of 60° with the positive Z-axis. And the arrow that is below the XY-plane, is making an obtuse angle of 120° with the positive Z-axis. 

If the third direction angle given with the positive Z-axis is 120°, then this confirms that the arrow below the XY plane is the given vector \(\vec{u}\). Also, this arrow will be one of the infinite arrows of lenghth 4 units from the origin along the lateral surface of the cone that form 120° angle with positive Z-axis. 

The intersection of three cones in the only line segment along which this arrow is drawn. 

So the conclusion in the end is that we require all three direction angles with positive X, Y and Z axes, to express the direction of a vector in 3D.