Problem on finding a vector perpendicular to a triangle

Consider the triangle in three-space given by $\langle a, 0, 0\rangle, \langle 0, b, 0 \rangle, \langle 0, 0, c \rangle.$ Find a vector that is perpendicular to the triangle and has length equal to the area of the triangle.
• Solution

Recall that
Hence, if we can find two vectors within the triangle, a cross product will get us close to the vector we seek.
We consider the vectors going from $\langle a, 0, 0\rangle$ to $\langle 0, b, 0\rangle$ or $\langle 0, 0, c\rangle.$
Recall that
The vectors from $\langle a, 0, 0\rangle$ to $\langle 0, b, 0\rangle$ or $\langle 0, 0, c\rangle$ are then $\langle -a, b, 0\rangle$ and $\langle -a, 0, c\rangle$:
Hence a vector perpendicular to the triangle is $$\langle -a, b, 0 \rangle \times \langle -a, 0, c \rangle.$$
We need to relate this vector to the length of the area of the triangle. So, we seek the relationship of a cross product to the area of some shape.
Recall that
We note that the parallelegram spanned by two vectors has twice the area as the triangle spanned by them:
Hence the perpendicular vector whose length equals the area of the triangle is: $$\frac{\langle -a, b, 0 \rangle \times \langle -a, 0, c\rangle}{2} =\left \langle \frac{bc}{2}, \frac{ac}{2}, \frac{ab}{2} \right \rangle.$$