Question:medium

Which of the following relations is true for two unit vector \(\hat A\) and \(\hat B\) making an angle θ to each other?

Updated On: Mar 23, 2026
  • \(|\hat A+\hat B| = |\hat A-\hat B|\ tan\frac {θ}{2}\)
  • \(|\hat A-\hat B| = |\hat A+\hat B|\ tan\frac {θ}{2}\)
  • \(|\hat A+\hat B| = |\hat A-\hat B|\ cos\frac {θ}{2}\)
  • \(|\hat A-\hat B| = |\hat A+\hat B|\ cos\frac {θ}{2}\)
Show Solution

The Correct Option is B

Solution and Explanation

To solve the given problem, let's analyze the mathematical relationships between two unit vectors \(\hat A\) and \(\hat B\) that make an angle \( \theta \) with each other. We are tasked with determining which relation is true among the provided options.

Let's start by understanding the properties of vectors:

  1. Since \(\hat A\) and \(\hat B\) are unit vectors, their magnitudes are 1: |\hat A| = 1 and |\hat B| = 1.
  2. The angle between the vectors is \(\theta\). Therefore, using the dot product property, we have: \(\hat A \cdot \hat B = \cos \theta\).

Now, let's calculate the magnitude of the vector sum and difference:

  1. The magnitude of |\hat A + \hat B| can be computed using the formula: |\hat A + \hat B| = \sqrt{(\hat A + \hat B) \cdot (\hat A + \hat B)} = \sqrt{|\hat A|^2 + |\hat B|^2 + 2(\hat A \cdot \hat B)} = \sqrt{2 + 2\cos \theta}.
  2. The magnitude of |\hat A - \hat B| is: |\hat A - \hat B| = \sqrt{(\hat A - \hat B) \cdot (\hat A - \hat B)} = \sqrt{|\hat A|^2 + |\hat B|^2 - 2(\hat A \cdot \hat B)} = \sqrt{2 - 2\cos \theta}.

Now, examine the correct relationship from the options:

Given the correct answer: |\hat A - \hat B| = |\hat A + \hat B| \cdot \tan\frac{\theta}{2}.

To verify this relation, recall that: \(\tan\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}\).

Thus, using the above expression for \(\tan\frac{\theta}{2}\) and substituting the derived magnitudes:

  1. Substitute \(|\hat A - \hat B|\): \(\sqrt{2 - 2\cos\theta}\)
  2. Substitute \(|\hat A + \hat B|\): \(\sqrt{2 + 2\cos\theta}\)
  3. Check your relation: \(\sqrt{2 - 2\cos\theta} = \sqrt{2 + 2\cos\theta} \cdot \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}\)

This verifies the relationship holds true with the given mathematical identities.

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