Question:medium

If \((\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 27\) and \(|\vec{a}| = 2|\vec{b}|\), then \(|\vec{b}|\) is:

Updated On: Jan 16, 2026
  • 3
  • 2
  • \( \frac{5}{6} \)
  • 6
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The Correct Option is A

Solution and Explanation

Given \( |\vec{a}| = 2|\vec{b}| \), we calculate \( |\vec{a} - \vec{b}| \) as follows:
The magnitude of the difference between two vectors is given by
\[ |\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}}. \]
From \( |\vec{a}| = 2|\vec{b}| \), it follows that \( |\vec{a}|^2 = 4|\vec{b}|^2 \).
Since \( \vec{a} \) and \( \vec{b} \) are in the same direction, their dot product is \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| = 2|\vec{b}|^2 \).
Substituting these into the magnitude formula yields:
\[ |\vec{a} - \vec{b}| = \sqrt{4|\vec{b}|^2 + |\vec{b}|^2 - 2 \cdot 2|\vec{b}|^2}. \]
Simplification leads to:
\[ |\vec{a} - \vec{b}| = \sqrt{4|\vec{b}|^2 + |\vec{b}|^2 - 4|\vec{b}|^2} = \sqrt{|\vec{b}|^2} = |\vec{b}|. \]
Given \( |\vec{b}| = 3 \), we find:
\[ |\vec{a} - \vec{b}| = 3. \]
The final result is: 3.

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