Question:
Two vectors of same magnitude act at a point. Twice the product of the magnitudes of two vectors is equal to the square of the magnitude of their resultant. The angle between the two vectors is:
Two vectors of same magnitude act at a point. Twice the product of the magnitudes of two vectors is equal to the square of the magnitude of their resultant. The angle between the two vectors is:
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For problems involving vectors, use the formula for the magnitude of the resultant vector and solve for the unknown angle.
Updated On: May 28, 2025
- \( 60^\circ \)
- \( 30^\circ \)
- \( 90^\circ \)
- \( 120^\circ \)
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The Correct Option is C
Approach Solution - 1
We are given that the two vectors are of the same magnitude. Let the magnitude of each vector be \( A \), and the angle between the two vectors be \( \theta \).
According to the problem, twice the product of the magnitudes of the two vectors is equal to the square of the magnitude of their resultant.
\[
2A^2 = R^2
\]
The magnitude of the resultant \( R \) can be calculated using the formula:
\[
R = \sqrt{A^2 + A^2 + 2A^2 \cos \theta} = \sqrt{2A^2(1 + \cos \theta)}
\]
We are told that:
\[
2A^2 = R^2
\]
Substituting the expression for \( R^2 \), we get:
\[
2A^2 = 2A^2(1 + \cos \theta)
\]
Simplifying:
\[
1 = 1 + \cos \theta
\]
\[
\cos \theta = 0
\]
\[
\theta = 90^\circ
\]
Thus, the angle between the two vectors is \( 90^\circ \).
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Approach Solution -2
Step 1: Let the magnitude of each vector be \( A \)
Given that both vectors have the same magnitude, let the magnitude be \( A \).
Step 2: Use the formula for the magnitude of the resultant of two vectors
If two vectors of magnitude \( A \) are inclined at angle \( \theta \), the magnitude of the resultant \( R \) is:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos\theta} = \sqrt{2A^2(1 + \cos\theta)} \]
So, \[ R^2 = 2A^2(1 + \cos\theta) \]
Step 3: Use the given condition
We are told that:
\[ 2 \cdot A \cdot A = R^2 \quad \Rightarrow \quad 2A^2 = R^2 \]
Step 4: Equate the expressions for \( R^2 \)
\[ 2A^2 = 2A^2(1 + \cos\theta) \quad \Rightarrow \quad 1 = 1 + \cos\theta \quad \Rightarrow \quad \cos\theta = 0 \]
Step 5: Solve for \( \theta \)
\[ \cos\theta = 0 \quad \Rightarrow \quad \theta = 90^\circ \]
Final Answer: The angle between the two vectors is \( 90^\circ \).
Given that both vectors have the same magnitude, let the magnitude be \( A \).
Step 2: Use the formula for the magnitude of the resultant of two vectors
If two vectors of magnitude \( A \) are inclined at angle \( \theta \), the magnitude of the resultant \( R \) is:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos\theta} = \sqrt{2A^2(1 + \cos\theta)} \]
So, \[ R^2 = 2A^2(1 + \cos\theta) \]
Step 3: Use the given condition
We are told that:
\[ 2 \cdot A \cdot A = R^2 \quad \Rightarrow \quad 2A^2 = R^2 \]
Step 4: Equate the expressions for \( R^2 \)
\[ 2A^2 = 2A^2(1 + \cos\theta) \quad \Rightarrow \quad 1 = 1 + \cos\theta \quad \Rightarrow \quad \cos\theta = 0 \]
Step 5: Solve for \( \theta \)
\[ \cos\theta = 0 \quad \Rightarrow \quad \theta = 90^\circ \]
Final Answer: The angle between the two vectors is \( 90^\circ \).
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