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This question arised from my other question How can I prove that 3 planes are arranged in a triangle-like shape without calculating their intersection lines?, where I asked how I can show that three vectors (the normals of three planes) lie in one plane, i.e. are linearly dependent.

One of the answers stated that "The three normals $n_1, n_2, n_3$ all lie in a plane $P$ through the origin, because $n_1 - n_2 = n_3.$". This makes sense to me and explained why I could intuitively solve the question given in school (I nevertheless accepted another answer because I felt that it provided a more complete solution to the problem).

Recently, I thought about the same problem again and wondered why exactly three vectors are linearly dependent if one can be formed by adding/subtracting the other two, i.e. one vector is a combination of the others. I would love to have both an intuitive answer and a mathematical proof (if possible, on highschool level).

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    $\begingroup$ There has been a downvote on this question. Could you please explain why you downvoted so that I can improve my question. Otherwise, downvoting is not helpful. $\endgroup$
    – jng224
    Commented Sep 15, 2020 at 17:09
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    $\begingroup$ I didn't down vote, and I agree that downvoting without explanation doesn't help you. Anyway, if $n_3=n_1-n_2$, then $n_1-n_2-n_3=0$, which meets my criterion for linear dependence ($a_1n_1+a_2n_2+a_3n_3=0$ with not all $a_1,a_2,a_3$ zero) $\endgroup$ Commented Sep 15, 2020 at 17:32

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It is trivial actually.Lets say $v_1$ ,$v_2$ and $v_3$ are vectors . If $v_1=v_2 + v_3$, then $v_2 + v_3 - v_1=0$, so there exists a solution for $c_1v_1+c_2v_2+c_3v_3=0$ which is different than $c_1=c_2=c_3=0$ and that means that they are linearly dependent

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    $\begingroup$ This also works when the third vector equals any other combination of the other two, for the same reasons. $\endgroup$
    – David K
    Commented Sep 17, 2020 at 16:44

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