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Step-by-Step Solution
Step 1: Express the magnitude conditions
We have two vectors of equal magnitudes, say each of magnitude $X$. The problem states that:
$|\overrightarrow{X} - \overrightarrow{Y}| = n \,|\overrightarrow{X} + \overrightarrow{Y}|$.
Let the angle between $\overrightarrow{X}$ and $\overrightarrow{Y}$ be $\theta$.
Step 2: Write the magnitudes in terms of $X$ and $\theta$
The magnitude squared of a sum or difference of two vectors of equal magnitude $X$ is given by the law of cosines:
$|\overrightarrow{X} - \overrightarrow{Y}|^2 = X^2 + X^2 - 2 \,X \,X \,\cos\theta = 2X^2 (1 - \cos\theta)$.
$|\overrightarrow{X} + \overrightarrow{Y}|^2 = X^2 + X^2 + 2 \,X \,X \,\cos\theta = 2X^2 (1 + \cos\theta)$.
Step 3: Incorporate the factor $n$ and square the relation
Since
$|\overrightarrow{X} - \overrightarrow{Y}| = n \,|\overrightarrow{X} + \overrightarrow{Y}|,$
we get:
$ \sqrt{2X^2 (1 - \cos\theta)}
= n \,\sqrt{2X^2 (1 + \cos\theta)}.
$
Squaring both sides:
$2X^2 (1 - \cos\theta)
= n^2 \,\bigl[\,2X^2 (1 + \cos\theta)\bigr].
$
Step 4: Simplify and solve for $\cos\theta$
Dividing throughout by $2X^2$ gives:
$1 - \cos\theta
= n^2 (1 + \cos\theta).
$
Rearrange terms to isolate $\cos\theta$:
$1 - \cos\theta
= n^2 + n^2 \cos\theta,
$
$1 - n^2
= \cos\theta (n^2 + 1),
$
$\cos\theta
= \frac{1 - n^2}{n^2 + 1}.
$
Step 5: Express the angle in the form given in the options
Notice that
$\displaystyle \frac{1 - n^2}{n^2 + 1}
= \frac{n^2 - 1}{-(n^2 + 1)}.
$
Thus, we can write:
$\theta = \cos^{-1}\Bigl(\frac{n^2 - 1}{-\,\bigl(n^2 + 1\bigr)}\Bigr).
$
This matches the form given as the correct answer.
Final Answer
$\displaystyle \theta
= \cos^{-1}\left(\frac{n^2 - 1}{-\bigl(n^2 + 1\bigr)}\right).
$
