Question

For any two vectors \( \vec{a} \) and \( \vec{b} \), prove that always \[ |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|. \]

Hint:

The Cauchy-Schwarz inequality provides an upper bound for the dot product of two vectors. It states that the absolute value of the dot product is less than or equal to the product of the magnitudes of the vectors.

Answer 1

We need to prove the inequality: \[ |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|. \] This is a result of the Cauchy-Schwarz inequality. The dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, \] where \( \theta \) is the angle between the two vectors.

Step 1: Analyze the cosine term. Since \( \cos \theta \) lies in the range \( -1 \leq \cos \theta \leq 1 \), we have: \[ |\cos \theta| \leq 1. \]

Step 2: Apply the inequality. Thus, the magnitude of the dot product is: \[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta| \leq |\vec{a}| |\vec{b}| \times 1. \] Conclusion: Therefore, we have shown that: \[ |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|. \]