Question

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitude of $(\vec{A} + \vec{B})$ is 'n' times the magnitude of $(\vec{A} - \vec{B})$ . The angle between $\vec{A}$ and $\vec{B}$ is :

• A. $\sin^{-1} \left[\frac{n^{2}-1}{n^{2}+1}\right] $
• B. $\cos^{-1} \left[\frac{n -1}{n+1}\right] $
• C. $\cos^{-1} \left[\frac{n^{2}-1}{n^{2}+1}\right] $
• D. $\sin^{-1} \left[\frac{n -1}{n+1}\right] $

Answer 1

The Correct Option is C

To solve this problem, let us start by understanding the given information:

1. We have two vectors, $\vec{A}$ and $\vec{B}$, with equal magnitudes. Let's denote the magnitude of each vector as $|\vec{A}| = |\vec{B}| = a$.
2. The magnitude of $(\vec{A} + \vec{B})$ is 'n' times the magnitude of $(\vec{A} - \vec{B})$. So, we have:

$$ |\vec{A} + \vec{B}| = n \cdot |\vec{A} - \vec{B}| $$

Let's express $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$ in terms of the magnitudes of vectors and the angle between them. Let the angle between $\vec{A}$ and $\vec{B}$ be $\theta$.

From vector addition and subtraction principles, we have:

• The magnitude of the sum of the vectors:

$$ |\vec{A} + \vec{B}| = \sqrt{a^2 + a^2 + 2a^2 \cos\theta} = \sqrt{2a^2(1 + \cos\theta)} $$

• The magnitude of the difference of the vectors:

$$ |\vec{A} - \vec{B}| = \sqrt{a^2 + a^2 - 2a^2 \cos\theta} = \sqrt{2a^2(1 - \cos\theta)} $$

Using the given condition:

$$ \sqrt{2a^2(1 + \cos\theta)} = n \cdot \sqrt{2a^2(1 - \cos\theta)} $$

We can simplify this equation as follows:

• Divide both sides by $\sqrt{2}a$:

$$ \sqrt{1 + \cos\theta} = n \cdot \sqrt{1 - \cos\theta} $$

• Square both sides:

$$ 1 + \cos\theta = n^2 (1 - \cos\theta) $$

• Expand and rearrange terms:

$$ 1 + \cos\theta = n^2 - n^2\cos\theta $$

• Combine terms involving $\cos\theta$:

$$ \cos\theta + n^2\cos\theta = n^2 - 1 $$

• Factor out $\cos\theta$:

$$ \cos\theta (1 + n^2) = n^2 - 1 $$

• Solve for $\cos\theta$:

$$ \cos\theta = \frac{n^2 - 1}{n^2 + 1} $$

Thus, the angle between $\vec{A}$ and $\vec{B}$ is:

$$ \theta = \cos^{-1} \left[\frac{n^2 - 1}{n^2 + 1}\right] $$

Therefore, the correct answer is: $\cos^{-1} \left[\frac{n^2 - 1}{n^2 + 1}\right] $.