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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?

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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 $|\vec b|=0$

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

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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 $\vec{b} = \vec{0}$.

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