The angle between vectors $\vec a$ and $\vec b$ is $120°$ and $|\vec a| = 3|\vec b|$. Define scalar λ so that $\vec a+\vec b$ and $\vec a-λ\vec b$ are perpendicular each other. The answer is $λ = −15$.
So far I have used this approach but don't get the right answer
$$(\vec a+\vec b)\cdot(\vec a-λ\vec b)=0 $$ $$\vec a^2-\vec aλ\vec b + \vec b\cdot\vec a - λ\vec b^2=0 $$ $$ λ= a/b $$
but don't know how to continue from this.
$$\vec a\cdot\vec a-\lambda\vec a\cdot\vec b+\vec a\cdot \vec b-\lambda\vec b\cdot\vec b=0$$ Thus $$|\vec a|^2+(1-\lambda)|\vec a||\vec b|\cos120^{\circ}-\lambda|\vec b|^2=0$$ Now put $|\vec a|=3|\vec b|$ $$9|\vec b|^2+3(1-\lambda) |\vec b|^2\left(-\frac12\right)-\lambda |\vec b|^2=0$$ If $|\vec b|≠0$, then $$9+3(1-\lambda) \left(-\frac12\right)-\lambda=0$$ from which $\lambda=-15$.