Question

For any two vectors \( \vec{a} \) and \( \vec{b} \), which of the following statements is always true ?

• A. \( \vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| \)
• B. \( |\vec{a} + \vec{b}| \geq |\vec{a}| + |\vec{b}| \)
• C. \( |\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}| \)
• D. \( |\vec{a} \times \vec{b}| \geq |\vec{a}| |\vec{b}| \)

Hint:

Cauchy-Schwarz (\( \vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| \)) and Triangle Inequality (\( |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \)) are the most important vector inequalities to memorize. They are fundamental to understanding vector projections and geometry.

Answer 1

The Correct Option is A

Concept:

• Scalar Product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.
• Properties of Cosine: The value of \( \cos \theta \) always lies in the range \( [-1, 1] \).
• Triangle Inequality: For any two vectors, \( |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \).

Step 1: Evaluate statement (A)
We know that \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \). Since \( \cos \theta \leq 1 \) for all \( \theta \):
\[ \vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| \cdot (1) \] \[ \vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| \]
This is the Cauchy-Schwarz inequality and it is always true.

Step 2: Evaluate statement (B)
According to the Triangle Inequality, \( |\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \).
The statement \( |\vec{a} + \vec{b}| \geq |\vec{a}| + |\vec{b}| \) is false except in the limiting case where vectors are parallel.

Step 3: Evaluate statement (D)
The magnitude of the cross product is \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).
Since \( \sin \theta \leq 1 \), it follows that \( |\vec{a} \times \vec{b}| \leq |\vec{a}| |\vec{b}| \).
Thus, statement (D) is false.