# Prove that if vectors a and b are both parallel and perpendicular, then at least one of a or b is the 0 vector.

How do I prove that if vectors a and b are both parallel and perpendicular then at least one of them is 0?

It seems intuitive that this should be true, but I'm having difficulty finding a proof. I know that 0 is perpendicular and parallel to every vector, and, intuitively, that it is the only such vector, but only intuitively.

Could anybody offer some help?

If $$\vec a \parallel \vec b$$, then $$\vec a\cdot\vec b=\pm|\vec a||\vec b|.$$

If $$\vec a \perp \vec b$$, then $$\vec a\cdot\vec b=0.$$

If both, then $$\pm|\vec a||\vec b|=0$$, so $$|\vec a|=0$$ and/or $$|\vec b|=0$$

• $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ Dec 24 '19 at 5:17
• Oh, of course! I’m not sure how I didn’t see this before — I feel quite silly. Thank you! Dec 24 '19 at 5:37

You could try proving it in 2 Dimensions and then extending the result to higher dimensions. Let $$A = a_1x +a_2y$$ and $$B = b_1x+b_2y$$

Now use $$A\cdot B = 0$$ and $$A*B=0$$ to get relations between the coefficients and ultimately the mathematical proof. Can you proceed?

If $$\vec{a} \parallel \vec{b}$$, then $$\vec{a} = c \vec{b}$$ or $$\vec{b} = c \vec{a}$$ for some scalar $$c$$. Without loss of generality, suppose $$\vec{a} = c \vec{b}$$.

If in addition, $$\vec{a} \perp \vec{b}$$ then $$\vec{a} \cdot \vec{b} = 0.$$

But then $$\vec{a} \cdot \vec{b} = c \vec{b} \cdot \vec{b} = c |\vec{b}|^2 = 0$$.

Hence either $$c = 0$$, which implies $$\vec{a} = 0 \vec{b} = \vec{0}$$, or $$|\vec{b}| = 0$$, which implies $$\vec{b} = \vec{0}$$.