Prove the following


If $\vec{x}$ and $\vec{y}$ be two non-zero vectors such that

$|\vec{x}+\vec{y}|=|\vec{x}|$ and $2 \vec{x}+\lambda \vec{y}$ is perpendicular to $\vec{y}$, then the value of $\lambda$ is_________.



Squaring both sides we get

$|\vec{x}|^{2}+2 \vec{x} \cdot \vec{y}+|\vec{y}|^{2}=|\vec{x}|^{2}$

$\Rightarrow 2 \vec{x} \cdot \vec{y}+\vec{y} \cdot \vec{y}=0$ $\ldots$ (i)

Also $2 \vec{x}+\lambda \vec{y}$ and $\vec{y}$ are perpendicular

$\therefore 2 \vec{x} \cdot \vec{y}+\lambda \vec{y} \cdot \vec{y}=0$ ...(ii)

Comparing (i) and (ii), $\lambda=1$

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