Question

Prove that for any two non-zero vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \). Also write the name of this inequality.

Hint:

Geometric intuition: In a triangle, the sum of any two sides is greater than or equal to the third side. Equality holds only if the vectors are collinear and in the same direction.

Answer 1

Step 1: Understanding the Concept:
The magnitude of the sum of two vectors is always less than or equal to the sum of their individual magnitudes.
Step 2: Key Formula or Approach:
Use the identity \( |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \).
Step 3: Detailed Explanation:
Expand the square of the magnitude:
\[ |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \] \[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \] By definition, \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta \). Since \( \cos \theta \leq 1 \):
\[ \vec{a} \cdot \vec{b} \leq |\vec{a}||\vec{b}| \] Therefore:
\[ |\vec{a} + \vec{b}|^2 \leq |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}| \] \[ |\vec{a} + \vec{b}|^2 \leq (|\vec{a}| + |\vec{b}|)^2 \] Taking the square root on both sides:
\[ |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \] This result is known as the Triangle Inequality.
Step 4: Final Answer:
The inequality is the Triangle Inequality.