Vectors perpendicular to triangle sides sum to zero

Question

I'm able to prove that, for a triangle $PQR$, $$ PQ+ QR+ RP = 0. $$ I am not sure how to show that for $A \perp PQ$, $B\perp QR$, and $C \perp RP$, where $A$ has the same length as $PQ$, $B$ has the same length as $QR$, and $C$ has the same length as $RP$, we have $$A+B+C=0.$$ I've made a sketch of the triangle and perpendicular vectors, but I'm not sure how to proceed.

Answer 1

I'll use a vector notation here, as in $\vec{PQ}+\vec{QR}+\vec{RP}=0$, which I hope is clear. (If not, please leave a comment and I'll alter the style of the answer.)

If you're familiar with the cross product, then if we treat these vectors as living in a three-dimensional space, we can take the above equation and take the cross-product with a unit vector $\hat{\vec{z}}$ in the third direction, giving, $$(\vec{PQ}+\vec{QR}+\vec{RP})\times \hat{\vec{z}}=0\,,$$ which is equivalent to $\vec{A}+\vec{B}+\vec{C}=0$. (Whether the perpendicular vectors point out of or into the triangle is just a matter of sign, and is determined by which way you let $\vec{PQ}$ etc point, or alternatively whether you choose 'out of' or 'into' the page as your third direction.)

Alternatively, we can think of it as follows. $\vec{PQ}+\vec{QR}+\vec{RP}=0$ is really two equations (one for the horizontal direction and one for the vertical direction). Putting the second equation in the 'top slot', and the first in the 'bottom slot' but with an added minus sign, (i.e. taking each vector $(\alpha,\beta)$ to $(\beta,-\alpha)$ in the equation) you get another correct vector equation, which is nothing but $\vec{A}+\vec{B}+\vec{C}=0$. (Unfortunately I don't know how to add reasonable diagrams, but you should be able to convince yourself that the vector $(b,-a)$ is perpendicular to the vector $(a,b)$, which is all that's required.)