If $\vec{a}$ and $\vec{b}$ makes an angle $\cos ^{-1}\left(\frac{5}{9}\right)$ with each other, then $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$ for $|\vec{a}|=n|\vec{b}|$ The integer value of $\mathrm{n}$ is ____

Question

Aicharya · Accepted Answer

$\begin{aligned} & \cos 	heta=\frac{5}{9} \ & \frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9} \ & a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b} \ & 6 \vec{a} \cdot \vec{b}=a^2+b^2\end{aligned}$ $\begin{aligned} & 6 	imes \frac{5}{9} a b=a^2+b^2 \ & \frac{10}{3} a b=a^2+b^2 \quad \& \quad a=n b \ & \frac{10}{3} n b^2=n^2 b^2+b^2 \ & 3 n^2-10 n+3=0 \ & n=\frac{1}{3} 	ext { and } n=3\end{aligned}$ integer value $n=3$

If $\vec{a}$ and $\vec{b}$ makes an angle $\cos ^{-1}\left(\frac{5}{9}\right)$ with each other, then…

Solution