Question

For two vectors \(\vec{a}\) and \(\vec{b}\) prove that \(|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|\).

Hint:

Proving vector identities and inequalities often involves starting with the square of the magnitude and using the identity \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\). This converts the problem from one of magnitudes (which involve square roots) to one of dot products, which are often easier to manipulate algebraically.

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. Geometrically, it means that the length of any side of a triangle is less than or equal to the sum of the lengths of the other two sides. The proof relies on the properties of the dot product and the Cauchy-Schwarz inequality.
Step 2: Key Formula or Approach:
1. Start with the square of the magnitude, \(|\vec{a} + \vec{b}|^2\).
2. Use the property \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\).
3. Expand the dot product \((\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})\).
4. Apply the Cauchy-Schwarz inequality, which states \(\vec{a} \cdot \vec{b} \le |\vec{a}| |\vec{b}|\).
5. Show that \(|\vec{a} + \vec{b}|^2 \le (|\vec{a}| + |\vec{b}|)^2\).
6. Take the square root of both sides to obtain the final inequality.
Step 3: Detailed Explanation:
We consider the square of the magnitude of \(\vec{a} + \vec{b}\): \[ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \] Expanding the dot product: \[ |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \] Using the properties \(\vec{v} \cdot \vec{v} = |\vec{v}|^2\) and \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\): \[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \] From the definition of the dot product, we know \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\), where \(\theta\) is the angle between the vectors. The Cauchy-Schwarz inequality states that \(\vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}|\) (since \(\cos\theta \le 1\)).
Applying this to our expression: \[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \le |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2 \] The right side of the inequality is the expansion of \((|\vec{a}| + |\vec{b}|)^2\).
So, we have: \[ |\vec{a} + \vec{b}|^2 \le (|\vec{a}| + |\vec{b}|)^2 \] Since magnitudes are non-negative, we can take the square root of both sides without changing the inequality: \[ |\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}| \] Step 4: Final Answer:
The triangle inequality for vectors, \(|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|\), has been proven. The equality holds when the vectors \(\vec{a}\) and \(\vec{b}\) are in the same direction (\(\theta = 0\)).