Let $a$ & $b$ be non-zero vectors such that $a · b = 0$. Use geometric description of scalar product to show that

Question

Let $a$ & $b$ be non-zero vectors such that $a \cdot b = 0$. Use geometric description of scalar product to show that $a$ & $b$ are perpendicular vectors.

What I've done so far is to state that $\cos 90$ (i.e. perpendicular) gives $0$ which would make the rest of the equation $|a||b|\cos \theta = 0$. I'm confused as to how set this out as a "formal proof"? Thanks in advance!

This is from the Cambridge Specialist 11 textbook.

Answer 1

You have all the correct ideas. It's now about putting them together in a coherent way. You are told that $a \cdot b= 0$. You know that $a \cdot b= |a| |b| \cos \theta$. Therefore, $0= |a| |b| \cos \theta$. So one of $|a|, |b|, \cos \theta$ is zero. Use what was given about $a,b$ to explain why $|a|,|b|$ are not zero. Then have you know, $\cos \theta = 0$. Now $\theta$ is the angle between $a,b$. Because $\cos \theta=0$, what are the possibilities for $\theta$? What does this mean about $a,b$?

Answer 2

You had it right.

It is given that $a$ and $b$ are non-zero, and the dot product is zero. This means that $$|a||b|\cos\theta=0$$

This means that one of the above factors is zero (null factor law). But we know that $a,b\neq0$, so the only way the above product is zero is if $\cos\theta=0$, which implies that $\theta=90°$, and hence the two vectors are perpendicular.