# Proofing the identity $(\vec{R}+\vec{a})\cdot(\vec{R}-\vec{a})=R^2-a^2$ graphically

From basic vector algebra, we know that $(\vec{R}+\vec{a})\cdot(\vec{R}-\vec{a})= R^2-\vec{R}\cdot\vec{a}+\vec{a}\cdot\vec{R}-a^2=R^2-a^2$, where the result is independent of the angle between $\vec{R}$ and $\vec{a}$.

It is trivial to prove using vector math. Are there any ways to prove this graphically, especially on the part that this expression does not depend on the angle?

Edit: No, finding the area spanned by the two vector won't help, because this is a dot product, not cross product. There doesn't seem to exist any visualizations of dot product, even after some googling.

• Try drawing it in 2 dimensions. Or three if you're feeling artistic – user238841 May 14 '16 at 2:57
• I tried that, but drawing a circle at the point of another vector doesn't seem to go anywhere. – user3664611 May 14 '16 at 3:02
• $\vec R+\vec a$ and $\vec R-\vec a$ are the diagonals of a parallelogram with sides $\vec R$ and $\vec a$. – amd May 14 '16 at 5:47
• Yes, I can see that. But how is it related to the dot product? Aren't parallelograms only relevant for cross products? – user3664611 May 14 '16 at 5:52