Why the Angle of the Sum of Two Vectors Isn't the Average of Their Individual Angles

The question seems to assume that each vector 'has an angle.' This only makes sense in (R^2) so I'll assume the question is about vectors in (R^2). In general, the angle (theta_{mathbf{a} mathbf{b}}) that the sum of two vectors (mathbf{a} mathbf{b}) makes with the (x)-axis is not the average of the angles formed separately by (mathbf{a}) and (mathbf{b}) with the (x)-axis. This happens only if either (mathbf{a}) and (mathbf{b}) have the same magnitude or they have the same direction.

A Diagrammatic Explanation

The easiest way to understand this is through the parallelogram law for vector addition. For any vector (mathbf{v}), the angle (theta_{mathbf{v}}) measures the angle made counterclockwise with the positive (x)-axis. The question incorrectly proposes that [theta_{mathbf{a} mathbf{b}} frac{theta_{mathbf{a}} theta_{mathbf{b}}}{2}] for all vectors (mathbf{a}) and (mathbf{b}) in (R^2).

This is already false for (mathbf{a} eq mathbf{0}) and (mathbf{b} mathbf{0}). Let's look a little deeper. Suppose we choose our labeling so that (theta_{mathbf{a}} theta_{mathbf{b}}). This assumes that (mathbf{a}) and (mathbf{b}) have different directions. Let (alpha) be the angle between (mathbf{a}) and (mathbf{a} mathbf{b}) and let (beta) be the angle between (mathbf{a} mathbf{b}) and (mathbf{b}). Then, we have

[theta_{mathbf{a} mathbf{b}} theta_{mathbf{a}} alpha quadtext{and}quad theta_{mathbf{b}} - theta_{mathbf{a}} alpha beta.]

Now, assume the incorrect statement proposed is true. Then, using both items in the above equations, we get

2theta_{mathbf{a} mathbf{b}} theta_{mathbf{a}} theta_{mathbf{b}} iff; 2theta_{mathbf{a}} 2alpha theta_{mathbf{a}} theta_{mathbf{b}} iff; theta_{mathbf{b}} - theta_{mathbf{a}} 2alpha iff; alphabeta 2alpha iff; alpha beta. end{aligned}]

For the above conditions to be true, we would like to invoke the parallelogram law for vector addition but in order to have a parallelogram, vectors (mathbf{a}) and (mathbf{b}) cannot have opposite directions.

For this special case, assuming (mathbf{a}) and (mathbf{b}) have different but not opposite directions, the fact that (alpha beta) indicates that the diagonal of the parallelogram bisects the vertex angles and this is true if and only if the parallelogram is a rhombus. But that means that if (mathbf{a}) and (mathbf{b}) have different directions, then (|mathbf{a}| |mathbf{b}|). Therefore, the proposed average relation for vectors with different but not opposite directions is true if and only if the two vectors have the same magnitude.

If (mathbf{a}) and (mathbf{b}) have the same direction, i.e., (theta_{mathbf{a}} theta_{mathbf{b}} theta_{mathbf{a} mathbf{b}}), then the statement is obviously true.

If (theta_{mathbf{b}} -theta_{mathbf{a}}), then their average is (mathbf{0}) and this doesn't have a defined angle with the x-axis.

If (theta_{mathbf{a}}) and (theta_{mathbf{b}}) have opposite directions but different magnitudes, then again, the statement does not hold.