A pyramid $OABC$ has vectors $\vec{OA}=a$, $\vec{OB}=b$ and $\vec{OC}=c$.

Question

A pyramid $OABC$ has vectors $\vec{OA}=a$, $\vec{OB}=b$ and $\vec{OC}=c$.

The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.

I have no idea where to even start with this so some help would be appreciated.

Answer 1

Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$ v_1 = \frac12 a \times b, $$ at least for one choice of orientation.

(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have $$ (-v_1) + v_2 + v_3 + v_4 = 0, $$ from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)

Answer 2

You have $$\vec v_1=\frac 12 \vec a\times \vec b$$ You have similar expressions for $\vec v_2$ and $\vec v_3$. The missing thing is $\vec v_4$. For that, you can see that that particular face is a triangle, with sides $\vec b-\vec a$ and $\vec c-\vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $\vec a\times\vec b$ pointing outwards. Similarly $\vec b\times\vec c$ and $\vec c\times\vec a$. Now the $\vec b-\vec a $ points from $A$ to $B$, and $\vec c -\vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $\vec b\times \vec a=-\vec a\times \vec b$.