# Prove A perpendicular to B if |a-b| = |a+b|

I'm absolutely exhausted working on this problem. I think I am very close, but I don't know what to do from here. I've tried a lot of different algebraic approaches.

So the problem says to prove A is perpendicular to B if |a-b| = |a+b|(we are talking about vectors and this is calculus class and this unit's recent lessons include the dot product and cross product).

What I've tried so far: I've tried drawing a diagram and using simple vector math and the property of a right triangle I noticed that if A and B are set up as connecting vectors at a right angle, then |a+b| would be the hypotenuse and |a-b| would be the hypotenuse of a mirrored right angle triangle. From there I tried plugging it into the Pythagorean formula as such:

C^2 = A^2 + B^2 since this would prove A and B are perpendicular Since C = |a+b| = |a-b| I substituted in...

|A+B||A-B| = |A|^2 + |B|^2

|A^2-B^2| = |A|^2 + |B|^2

I've been working on this for hours trying rearranging and I think I am need to try something new. I realize the dot product will prove perpendicular, but I don't know how to link the two. I'm stumped;

Any help is appreciated.

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If $|a-b|=|a+b|$ then $|a-b|^2=|a+b|^2$. Use the fact that $|v|^2=v\cdot v$. – wj32 Apr 18 '13 at 2:11

$$\|\vec{a} + \vec{b}\| = \|\vec{a} - \vec{b}\|$$ Dotting a vector with itself is the square of its length: $$\|\vec{a} + \vec{b}\|^2 = \|\vec{a} - \vec{b}\|^2$$ $$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$$ Dot product distributes over addition: $$\vec{a} \cdot (\vec{a} + \vec{b}) + \vec{b} \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot (\vec{a} - \vec{b}) - \vec{b} \cdot (\vec{a} - \vec{b})$$ $$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$$ $$2 (\vec{a} \cdot \vec{b}) = -2 (\vec{a} \cdot \vec{b})$$ $$\vec{a} \cdot \vec{b} = 0$$ Vectors dot to $0$ iff they are perpendicular: $\vec{a}$ is perpendicular to $\vec{b}$.

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I see... the only way that 2(a.b) can equal -2(a.b) is if (a.b) = 0. Thank you for a great answer, I should've asked for help sooner. – Klik Apr 18 '13 at 2:41

I assume you are working with real inner products. Since $\|a-b\|=\|a+b\|$, we have $\langle a-b,a-b\rangle=\langle a+b,a+b\rangle$. But $$\langle a-b,a-b\rangle = \langle a,a-b\rangle - \langle b,a-b\rangle =\langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle$$ and $$\langle a+b,a+b\rangle = \langle a,a+b\rangle + \langle b,a+b\rangle =\langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle$$ so subtracting the two right-hand sides gives us $\langle a,b\rangle + \langle b,a\rangle=0$. What can you conclude from this?

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I just learned a lot of new notations from reading your post, thanks for that. I appreciate the help. And yes... the only way 2<a,b> = 0 is if <a,b> = 0 and therefore perpendicular. :) – Klik Apr 18 '13 at 2:56
If you are confused by the notation: $\langle a,b \rangle$ is much more general than $a \cdot b$. It can denote any kind of inner product. It shows up in geometry (dot product), quantum mechanics (bra-ket notation), statistics (expected value), etc. – Henry Swanson Apr 18 '13 at 3:14

Although the vector algebra in the other answers is probably what was intended when the problem was posed, I can't resist pointing out that, as you might have noticed when you drew pictures, the problem amounts to the theorem, from Euclidean geometry, that if the diagonals of a parallelogram have the same length then the parallelogram is a rectangle.

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I'm really glad you are commenting because I was wondering why my attempt with using the diagram did not work. Can you elaborate on this, or provide a link explaining this more? I'm not familiar with the theorem you're speaking of. Have a look at my illustration and let me know what you think. img221.imageshack.us/img221/6893/conceptab.png – Klik Apr 18 '13 at 3:48