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

\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: $n = 3$.