Physics - Vectors Question with Solution | TestHub

Statement-1: We know that $|\vec{A}+\vec{B}{|}^2=(\vec{A}+\vec{B})\cdot (\vec{A}+\vec{B})=|\vec{A}{|}^2+|\vec{B}{|}^2+2\vec{A}\cdot \vec{B}$. Similarly, $|\vec{A}-\vec{B}{|}^2=|\vec{A}{|}^2+|\vec{B}{|}^2-2\vec{A}\cdot \vec{B}$. Adding these two equations gives $|\vec{A}+\vec{B}{|}^2+|\vec{A}-\vec{B}{|}^2=2(|\vec{A}{|}^2+|\vec{B}{|}^2)$. Thus, Statement-1 is true.

Statement-2: If $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$, then squaring both sides yields $|\vec{A}+\vec{B}{|}^2=|\vec{A}-\vec{B}{|}^2$.$$|\vec{A}{|}^2+|\vec{B}{|}^2+2\vec{A}\cdot \vec{B}=|\vec{A}{|}^2+|\vec{B}{|}^2-2\vec{A}\cdot \vec{B}$$ $$4\vec{A}\cdot \vec{B}=0⟹\vec{A}\cdot \vec{B}=0$$ This implies that $\vec{A}$ and $\vec{B}$ are perpendicular (assuming they are non-zero vectors). Thus, Statement-2 is true.

Statement-2 describes a condition for perpendicularity, which is an independent property from the parallelogram law stated in Statement-1. Hence, Statement-2 is not a correct explanation for Statement-1.