Question

For the two vectors \( \vec{a} \) and \( \vec{b} \), prove that \( |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \) when \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \).

Hint:

The key to proving vector magnitude inequalities is often to square both sides. This allows you to convert the problem into one involving dot products, which have convenient algebraic properties.

Answer 1

Step 1: Understanding the Concept:
This inequality is known as the triangle inequality for vectors. It states that the magnitude of the sum of two vectors is less than or equal to the sum of their individual magnitudes. We can prove it using the properties of the dot product.
Step 2: Key Approach:
It is easier to work with the squares of the magnitudes to avoid dealing with square roots. We will prove the equivalent inequality \( |\vec{a} + \vec{b}|^2 \leq (|\vec{a}| + |\vec{b}|)^2 \).
Step 3: Detailed Proof:
We start by considering the square of the magnitude of the sum of the vectors: \[ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \] Expanding the dot product using its distributive property: \[ = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \] We know that \( \vec{v} \cdot \vec{v} = |\vec{v}|^2 \) and the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)). \[ = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \] By the definition of the dot product, \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.
We know that the maximum value of \( \cos \theta \) is 1. Therefore, we have the inequality: \[ \vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| \] Substituting this into our expression: \[ |\vec{a} + \vec{b}|^2 \leq |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2 \] The right side of the inequality is the expansion of a perfect square: \[ |\vec{a} + \vec{b}|^2 \leq (|\vec{a}| + |\vec{b}|)^2 \] Since magnitude is a non-negative quantity, we can take the square root of both sides without changing the direction of the inequality: \[ |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \] Step 4: Final Answer:
We have successfully proven the triangle inequality for vectors.