Question

Prove that \(|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|\) is always true for any two vectors \(\vec{a}\) and \(\vec{b}\).

Hint:

This is a fundamental vector inequality. Remembering the proof, which simply relies on the definition of the dot product and the fact that \(|\cos\theta| \le 1\), is very useful for vector-based problems in physics and mathematics.

Answer 1

Step 1: Understanding the Concept:
The inequality \(|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|\) is known as the Cauchy-Schwarz inequality for vectors in Euclidean space. It relates the dot product of two vectors to the product of their magnitudes.
Step 2: Key Formula or Approach:
The proof relies on the geometric definition of the dot product (scalar product) of two vectors \(\vec{a}\) and \(\vec{b}\):
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \] where \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them.
Step 3: Detailed Explanation or Calculation:
We start with the definition of the dot product: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \] Take the absolute value (or modulus) of both sides of the equation: \[ |\vec{a} \cdot \vec{b}| = \left| |\vec{a}| |\vec{b}| \cos\theta \right| \] Since the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\) are non-negative scalars, we can take them out of the absolute value sign: \[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos\theta| \] Now, we consider the property of the cosine function. For any angle \(\theta\), the value of \(\cos\theta\) lies in the range \([-1, 1]\).
This means that the absolute value of \(\cos\theta\) must be in the range \([0, 1]\).
\[ 0 \le |\cos\theta| \le 1 \] Since \(|\cos\theta| \le 1\), we can multiply both sides of this inequality by the non-negative quantity \(|\vec{a}| |\vec{b}|\) without changing the inequality direction: \[ |\vec{a}| |\vec{b}| |\cos\theta| \le |\vec{a}| |\vec{b}| \times 1 \] Substituting \(|\vec{a} \cdot \vec{b}|\) back in place of \(|\vec{a}| |\vec{b}| |\cos\theta|\), we get: \[ |\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}| \] Step 4: Final Answer:
The inequality is proven based on the fundamental properties of the dot product and the cosine function. The equality holds when \(|\cos\theta|=1\), which means \(\theta = 0\) or \(\theta = \pi\) (i.e., when the vectors are collinear).