Proof with 3D vectors Let ${a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, ${b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and ${c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
${a} \times {b} + {b} \times {c} + {c} \times {a} ={0}.$

Hello,

Is there any other way to do this problem without bashing?  I can't seem to find a nice and slick solution to this.

Thanks in advance!
The points are collinear iff the area of the triangle bounded by the points is zero.

$$\text{ Area } = ||(\vec a - \vec b ) \times (\vec a - \vec c ) ||$$

$$ = ||\vec a \times \vec a  - \vec a \times \vec c -  \vec b \times \vec a + \vec b \times \vec c ) ||   $$

$$ = ||0 + \vec c \times \vec a +  \vec a \times \vec b + \vec b \times \vec c ) ||   $$
In order to complete an "if and only if" proof, what you need to do is prove the statement in both directions. What you must prove is,

1) If $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear, then $a\times b + b\times c + c \times a = 0$.

2) If $a\times b + b\times c + c \times a = 0$, then $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear.

The way you start is by automatically assuming the "If" part is true. So for part 1), you're going to start by assuming that $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear. For part 2), you're going to assume that $a\times b + b\times c + c \times a = 0$. You don't need to prove $those$ parts because you're going to assume they're already true. What you must prove is the part after "then".