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Say we have a closed loop system of vectors, of which we know the magnitudes:

$$ \vec{OA} + \vec{AB} = \vec{BO}\\ |\vec{OA}| = 7200\\ |\vec{AB}| = 3600\\ |\vec{BO}| = 4900\\ $$

We also know that $\vec{OA}$ only has an $\hat{i}$ component.

The problem requires we find vector $\vec{AB}$ and thus $\vec{BO}$

This came up in an engineering problem I'm trying to solve. I know how to do it graphically. One only has to draw arcs at radii scaled to the magnitudes from their corresponding origins, and the point of intersection is point B. But I was curious how this is done analytically.

Cheers

Andy
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  • found an answer:

    https://math.stackexchange.com/questions/543961/determine-third-point-of-triangle-when-two-points-and-all-sides-are-known

     – Andy May 26 '18 at 11:30

1 Answers1

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We have $\vec{OA}=7200\hat i$. For a closed polygon, the sum of the ordered vectors for the sides is $0$. $\vec{OA}+\vec{AB}+\vec{BO}=0$. We can decompose $\vec{AB}$ into components $$\vec{AB}=x\hat i+y\hat j$$ I am skiping a step where I use the sum of the vectors, to write $$\vec{BO}=-(7200+x)\hat i-y\hat j$$If you write the equations for the magnitudes of those vectors, you get $$x^2+y^2=3600^2\\(7200+x^2)+y^2=4900^2$$ Subtracting the first equation from the second, you get a linear equation in $x$. Plug this value into the first equation and you get $y=\pm\sqrt{3600^2-x^2}$, meaning that you can draw two identical triangles, with $B$ on either side of $OA$.

Andrei
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