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Step-by-Step Solution
Step 1: Define the magnitudes of the vectors
Let the magnitudes of both vectors be equal to $x$, i.e.,
$| \overrightarrow{P} | = | \overrightarrow{Q} | = x$.
Step 2: Express the given condition
We are given that the magnitude of $ \overrightarrow{P} + \overrightarrow{Q} $ is $n$ times the magnitude of $ \overrightarrow{P} - \overrightarrow{Q} $:
$|\overrightarrow{P} + \overrightarrow{Q}| = n \,|\overrightarrow{P} - \overrightarrow{Q}|.$
Step 3: Square both sides to simplify
Squaring both sides, we get
$|\overrightarrow{P} + \overrightarrow{Q}|^2 \;=\; n^2 \, |\overrightarrow{P} - \overrightarrow{Q}|^2.$
Step 4: Use the formula for the square of the magnitude of a sum/difference of two vectors
Recall that for two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$,
$|\overrightarrow{A} + \overrightarrow{B}|^2 = A^2 + B^2 + 2AB \cos \theta,$
$|\overrightarrow{A} - \overrightarrow{B}|^2 = A^2 + B^2 - 2AB \cos \theta,$
where $A = |\overrightarrow{A}|$, $B = |\overrightarrow{B}|$, and $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.
Step 5: Substitute into our problem
Let $\theta$ be the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$. Then
$|\overrightarrow{P} + \overrightarrow{Q}|^2 = P^2 + Q^2 + 2PQ \cos \theta,$
$|\overrightarrow{P} - \overrightarrow{Q}|^2 = P^2 + Q^2 - 2PQ \cos \theta.$
Given $P = Q = x$, we have:
$(|\overrightarrow{P} + \overrightarrow{Q}|)^2 = x^2 + x^2 + 2x \cdot x \cos \theta = 2x^2 (1 + \cos \theta),$
$(|\overrightarrow{P} - \overrightarrow{Q}|)^2 = x^2 + x^2 - 2x \cdot x \cos \theta = 2x^2 (1 - \cos \theta).$
Step 6: Apply the condition and simplify for cos θ
From the given condition:
$2x^2 (1 + \cos \theta) = n^2 \,[\,2x^2 (1 - \cos \theta)\,].$
Cancelling $2x^2$ from both sides gives:
$1 + \cos \theta = n^2 (1 - \cos \theta).$
Expand and rearrange:
$1 + \cos \theta = n^2 - n^2 \cos \theta.$
Bring all cosine terms to one side:
$\cos \theta + n^2 \cos \theta = n^2 - 1.$
Factor out $\cos \theta$:
$\cos \theta (1 + n^2) = n^2 - 1.$
Hence,
$\cos \theta = \frac{n^2 - 1}{n^2 + 1}.$
Step 7: Final expression for the angle
Therefore, the angle $\theta$ between the vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is
$\theta = \cos^{-1}\!\Bigl(\frac{n^2 - 1}{n^2 + 1}\Bigr).$
This matches Option 4 from the given choices.
