The Triangle Inequality in Vector Addition

Problem: Triangle Inequality in Vector Addition

In classical mechanics and vector algebra, physical quantities such as displacement, velocity, acceleration, and force are represented by vectors. Let ##\vec{A}## and ##\vec{B}## be two vectors in a Euclidean vector space. Their resultant vector is defined as:

###\vec{R}=\vec{A}+\vec{B}###

Prove, using vector algebra and the dot product, that the magnitude of the resultant vector satisfies the triangle inequality:

###\left|\left|\vec{A}\right|-\left|\vec{B}\right|\right| \le \left|\vec{A}+\vec{B}\right| \le \left|\vec{A}\right|+\left|\vec{B}\right|###

Also analyze the physical and geometric conditions under which equality holds.

Worked Solution and Step-by-Step Explanation

To prove the result clearly, use the following notation:

###A=\left|\vec{A}\right|,\qquad B=\left|\vec{B}\right|,\qquad R=\left|\vec{R}\right|=\left|\vec{A}+\vec{B}\right|###

Here, ##A##, ##B##, and ##R## are non-negative scalar magnitudes. This distinction is important because ##\vec{A}## and ##\vec{B}## are vectors, while ##A## and ##B## are their scalar lengths.

Step 1: Express the Resultant Using the Dot Product

Since the magnitude squared of any vector is equal to its dot product with itself, we write:

###R^2=\vec{R}\cdot\vec{R}###

Using ##\vec{R}=\vec{A}+\vec{B}##:

###R^2=(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})###

Expanding the dot product:

###R^2=\vec{A}\cdot\vec{A}+\vec{A}\cdot\vec{B}+\vec{B}\cdot\vec{A}+\vec{B}\cdot\vec{B}###

Since the dot product is commutative,

###\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}###

and since

###\vec{A}\cdot\vec{A}=A^2,\qquad \vec{B}\cdot\vec{B}=B^2###

we obtain:

###R^2=A^2+B^2+2(\vec{A}\cdot\vec{B})###

If ##\theta## is the angle between ##\vec{A}## and ##\vec{B}##, then:

###\vec{A}\cdot\vec{B}=AB\cos\theta###

Therefore:

###R^2=A^2+B^2+2AB\cos\theta###

This is the standard magnitude formula for the resultant of two vectors.

Step 2: Proving the Upper Bound

For every real angle ##\theta##,

###-1\le \cos\theta \le 1###

To obtain the maximum possible value of ##R##, use ##\cos\theta \le 1##:

###R^2=A^2+B^2+2AB\cos\theta \le A^2+B^2+2AB###

The right-hand side is a perfect square:

###A^2+B^2+2AB=(A+B)^2###

Hence:

###R^2\le (A+B)^2###

Since ##R##, ##A##, and ##B## are non-negative, we can safely take the principal square root of both sides:

###R\le A+B###

Substituting back the meanings of ##R##, ##A##, and ##B##:

###\left|\vec{A}+\vec{B}\right|\le \left|\vec{A}\right|+\left|\vec{B}\right|###

This proves the upper bound of the triangle inequality.

Step 3: Proving the Lower Bound

To obtain the minimum possible value of ##R##, use ##\cos\theta \ge -1##:

###R^2=A^2+B^2+2AB\cos\theta \ge A^2+B^2-2AB###

The right-hand side is also a perfect square:

###A^2+B^2-2AB=(A-B)^2###

Therefore:

###R^2\ge (A-B)^2###

Taking the principal square root requires care. The quantity ##A-B## may be positive, negative, or zero. Therefore:

###\sqrt{(A-B)^2}=\left|A-B\right|###

Hence:

###R\ge \left|A-B\right|###

Substituting back the vector magnitudes:

###\left|\vec{A}+\vec{B}\right|\ge \left|\left|\vec{A}\right|-\left|\vec{B}\right|\right|###

This proves the lower bound.

Step 4: Combining the Two Bounds

From Step 2:

###\left|\vec{A}+\vec{B}\right|\le \left|\vec{A}\right|+\left|\vec{B}\right|###

From Step 3:

###\left|\vec{A}+\vec{B}\right|\ge \left|\left|\vec{A}\right|-\left|\vec{B}\right|\right|###

Combining both results:

###\boxed{\left|\left|\vec{A}\right|-\left|\vec{B}\right|\right| \le \left|\vec{A}+\vec{B}\right| \le \left|\vec{A}\right|+\left|\vec{B}\right|}###

This is the triangle inequality for vector addition.

Geometric Interpretation

Vector addition can be represented geometrically by placing the tail of ##\vec{B}## at the head of ##\vec{A}##. The resultant vector ##\vec{R}=\vec{A}+\vec{B}## joins the initial point of ##\vec{A}## to the final point of ##\vec{B}##.

This forms a triangle whose side lengths are ##A##, ##B##, and ##R##. In any triangle, one side cannot be greater than the sum of the other two sides. Therefore:

###R\le A+B###

Similarly, one side cannot be smaller than the absolute difference of the other two sides. Therefore:

###R\ge |A-B|###

Thus, geometry gives the same result:

###|A-B|\le R\le A+B###

Equality Conditions

The equality cases occur when the two vectors are collinear, meaning they lie along the same straight line.

1. Upper Equality Case

The upper equality

###\left|\vec{A}+\vec{B}\right|=\left|\vec{A}\right|+\left|\vec{B}\right|###

holds when ##\vec{A}## and ##\vec{B}## point in the same direction. In this case:

###\theta=0,\qquad \cos\theta=1###

Then:

###R^2=A^2+B^2+2AB=(A+B)^2###

So:

###R=A+B###

Physically, this means that two vector effects reinforce each other completely. For example, if two forces act on an object in the same direction, the resultant force is the sum of their magnitudes.

2. Lower Equality Case

The lower equality

###\left|\vec{A}+\vec{B}\right|=\left|\left|\vec{A}\right|-\left|\vec{B}\right|\right|###

holds when ##\vec{A}## and ##\vec{B}## point in exactly opposite directions. In this case:

###\theta=\pi,\qquad \cos\theta=-1###

Then:

###R^2=A^2+B^2-2AB=(A-B)^2###

So:

###R=|A-B|###

Physically, this means that the two vector effects oppose each other. For example, if two forces act in opposite directions, the resultant force is the difference between their magnitudes.

3.Special Case: Zero Vector

If either ##\vec{A}## or ##\vec{B}## is the zero vector, then both the upper and lower bounds become equal. For example, if ##\vec{B}=\vec{0}##, then:

###\left|\vec{A}+\vec{0}\right|=\left|\vec{A}\right|###

and the inequality becomes:

###\left|\vec{A}\right|\le \left|\vec{A}\right|\le \left|\vec{A}\right|###

So equality holds throughout.

Final Conclusion

For any two vectors ##\vec{A}## and ##\vec{B}## in a Euclidean vector space, the magnitude of their resultant vector ##\vec{A}+\vec{B}## is always bounded between the absolute difference and the sum of their individual magnitudes:

###\boxed{\left|\left|\vec{A}\right|-\left|\vec{B}\right|\right| \le \left|\vec{A}+\vec{B}\right| \le \left|\vec{A}\right|+\left|\vec{B}\right|}###

The upper limit occurs when the vectors point in the same direction. The lower limit occurs when the vectors point in exactly opposite directions. In all other cases, the resultant magnitude lies strictly between these two limits.