Question

Show that for any two vectors \( \mathbf{a} \) and \( \mathbf{b} \), always \( | \mathbf{a} \cdot \mathbf{b} | \leq \|\mathbf{a}\| \|\mathbf{b}\| \).

Hint:

The Cauchy-Schwarz inequality gives an upper bound for the absolute value of the dot product in terms of the magnitudes of the vectors.

Answer 1

Step 1: Understanding the dot product.
The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by the formula: \[ \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta, \] where \( \theta \) is the angle between the two vectors.

Step 2: Apply the Cauchy-Schwarz inequality.
By the Cauchy-Schwarz inequality, we know: \[ | \mathbf{a} \cdot \mathbf{b} | \leq \|\mathbf{a}\| \|\mathbf{b}\|. \] This is because \( \cos \theta \) lies in the range \( [-1, 1] \), so the magnitude of the dot product can never exceed the product of the magnitudes of the vectors.

Step 3: Conclusion.
Thus, we have shown that for any two vectors \( \mathbf{a} \) and \( \mathbf{b} \), it always holds that: \[ | \mathbf{a} \cdot \mathbf{b} | \leq \|\mathbf{a}\| \|\mathbf{b}\|. \]