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

Question

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

Professor PC Thomas Classes · Answer

(3) : $\cos 	heta=5 / 9$\ or $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$\ $a^{2}+b^{2}+2 \vec{a} \cdot \vec{b}=(\sqrt{2})^{2}\left(a^{2}+b^{2}-2 \vec{a} \cdot \vec{b}ight)$\ or $6 \vec{a} \cdot \vec{b}=a^{2}+b^{2}$\ or $6 	imes \frac{5}{9} a b=a^{2}+b^{2} \quad(\because \vec{a} \cdot \vec{b}=a b \cos 	heta)$\ or $2 	imes \frac{5}{3} \cdot n b^{2}=n^{2} b^{2}+b^{2}$\ $(\because a=n b($ given $))$\ or $3 n^{2}-10 n+3=0$ or $n=3$