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Suppose that $A$, $B$ and $C$ are non-zero vectors in $\mathbb R^2$. Show that if $A$ and $B$ are orthogonal and $B$ and $C$ are orthogonal then $A$ and $C$ are parallel.

I feel like this should be very simple. My initial thought was to do a proof by contradiction. So if we assume that $A$ is no parallel to $C$, then we end up with two different $B$'s, one orthogonal to $A$ and one to $C$, but I don't think that is rigorous enough.

User123454321
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  • Tbh i dont think the statement is true , it is true in 2 dimensions but certainly not in 3 , take A in x B in y and C in z direction , as you see A and C are not parallel – avz2611 May 10 '16 at 05:15
  • Sorry, I should have mentioned that. I tagged it as Euclidean Geometry meaning that it is in $\mathbb R^2$ – User123454321 May 10 '16 at 05:16

2 Answers2

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Since $A \perp B$, we have $B^TA=0$ implying $A$ and $B$ as two linearly independent vectors of $R^2$. Thus $C=\lambda_1A+\lambda_2B$ for some $\lambda_1,\lambda_2\in R$. Since $B\perp C$, $B^TC=0$

But $$B^TC=\lambda_1B^TA+\lambda_2B^TB$$ Hence $$\lambda_2B^TB=0$$ i.e. $\lambda_2=0$ (since $B\ne0$). So $$C=\lambda_1A$$ implying $$A \parallel C$$

Sry
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perpendicular$\Rightarrow$product of slope$=-1$.

Slope of $A$=Slope of $C$=$-1$/(Slope of $B$)

so they are parallel.