Question:

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

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For any two vectors, the absolute value of their dot product is always bounded by the product of their magnitudes, i.e., \( |\vec{a} \cdot \vec{b}| \leq |\vec{a}| \, |\vec{b}| \).
Updated On: Jan 27, 2025
  • \( \vec{a} \cdot \vec{b} \geq |\vec{a}| \, |\vec{b}| \)
  • \( \vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \)
  • \( \vec{a} \cdot \vec{b} \leq |\vec{a}| \, |\vec{b}| \)
  • \( \vec{a} \cdot \vec{b} \geq -|\vec{a}| \, |\vec{b}| \)
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The Correct Option is C

Solution and Explanation

Step 1: Definition of the dot product.
The dot product between two vectors \( \vec{a} \) and \( \vec{b} \) is defined as: \[ \vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \cos \theta, \] where \( \theta \) represents the angle between the vectors \( \vec{a} \) and \( \vec{b} \). Step 2: Investigate the range of \( \cos \theta \).
Since \( \cos \theta \) lies within the range \( -1 \leq \cos \theta \leq 1 \), it follows that: \[ -|\vec{a}| \, |\vec{b}| \leq \vec{a} \cdot \vec{b} \leq |\vec{a}| \, |\vec{b}|. \] Step 3: Final Answer.
Thus, the correct inequality for the dot product is: \[ \boxed{\vec{a} \cdot \vec{b} \leq |\vec{a}| \, |\vec{b}|}. \]
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