Question

If $\vec{a}$ and $\vec{b}$ make an angle $\cos^{-1}\left(\frac{5}{9}\right)$ with each other, then \[ |\vec{a} + \vec{b}| = \sqrt{2} |\vec{a} - \vec{b}| \quad \text{for } |\vec{a}| = n |\vec{b}|. \] The integer value of $n$ is _____.

Answer 1

Correct Answer: 3

\[\cos \theta = \frac{5}{9}\]\[\frac{\vec{a} \cdot \vec{b}}{ab} = \frac{5}{9} \hspace{20pt}(1)\]

 Using \(|\vec{a} + \vec{b}|^2 = \sqrt{2}|\vec{a} - \vec{b}|^2\): 
\[ a^2 + b^2 + 2a \cdot b = 2(a^2 + b^2 - 2a \cdot b) \]
Simplify: 
\[ a^2 + b^2 + 2a \cdot b = 2a^2 + 2b^2 - 4a \cdot b \] \[ 6(a \cdot b) = a^2 + b^2  \hspace{20pt}(2)\]

Substitute \(\vec{a} \cdot \vec{b} = ab \cdot \frac{5}{9}\) from (1):
\[ 6 \left( \frac{5}{9} ab \right) = a^2 + b^2 \]

Assume \(a = nb\): \[ \frac{10}{3} ab = a^2 + b^2 \]
\[ \frac{10}{3} nb^2 = n^2b^2 + b^2 \]
Divide through by \(b^2\): \[ \frac{10}{3}n = n^2 + 1 \]
Rearrange: \[ 3n^2 - 10n + 3 = 0 \]

Solve using the quadratic formula:
\[ n = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(3)}}{2(3)} \]
\[ n = \frac{10 \pm \sqrt{100 - 36}}{6} \] \[ n = \frac{10 \pm 8}{6} \]
\[ n = \frac{18}{6} = 3 \] (only positive integer value).

Final Answer: \(n = 3\).