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