Let a, b, c be mutually perpendicular, equal magnitude.

Question

Let $\vec a,\vec b,\vec c$ be mutually perpendicular, equal magnitude. If $\vec r$ satisfies
$$\vec a\times\big[(\vec r-\vec b)\times\vec a\big]+\vec b\times\big[(\vec r-\vec c)\times\vec b\big]+\vec c\times\big[(\vec r-\vec a)\times\vec c\big]=\vec 0$$
then $\vec r$ equals:

Accepted Answer

Using $\vec X\times(\vec Y\times\vec X) = |\vec X|^2\vec Y - (\vec X\cdot\vec Y)\vec X$ on each term, and $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$:

Since all magnitudes equal $|\vec a|$:
$$|\vec a|^2\big[3\vec r-(\vec a+\vec b+\vec c)\big] - \big[(\vec r\cdot\vec a)\vec a+(\vec r\cdot\vec b)\vec b+(\vec r\cdot\vec c)\vec c\big] = \vec 0$$

The bracketed term equals $|\vec a|^2\vec r$ (resolving $\vec r$ along the mutually perpendicular basis), giving $|\vec a|^2[3\vec r-(\vec a+\vec b+\vec c)-\vec r]=\vec 0$:
$$2\vec r = \vec a+\vec b+\vec c \implies \vec r = \frac12(\vec a+\vec b+\vec c)$$

Solutions · 1