cross productderivativesproof-explanationvectors
Using the vector cross product, how would I derive a formula for the area of a triangle with vertices: $$\\(x_0, y_0, z_0)\\(x_1, y_1, z_1)\\(x_2, y_2, z_2) $$ in terms of only $x_0, y_0, z_0, x_1, y_1, z_1, x_2, y_2, z_2. $
I would appreciate any help given, thank you very much!
You've defined $z_{1} \cdot z_{2} = Re(\overline{z_{1}} z_{2})$, and you want to show that $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$. Without loss of generality, assume $\arg(z_{1}) \geq \arg(z_{2})$, and note that $z_{1} \perp z_{2}$ if, and only if, $\arg(z_{1}) - \arg(z_{2}) = \frac{\pi}{2}$ or $\arg(z_{1}) - \arg(z_{2}) = \frac{3\pi}{2}$.
Next, write the two complex numbers in polar form: $z_{1} = |z_{1}| e^{i \arg(z_{1})}$ and $z_{2} = |z_{2}| e^{i \arg(z_{2})}$. Then we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(-\arg(z_{1}))\cos(\arg(z_{2}) - |z_{1}| |z_{2}| \sin(-\arg(z_{1})) \sin(\arg(z_{2})).$$ Because cosine is even and sine is odd, we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(\arg(z_{1}))\cos(\arg(z_{2}) + |z_{1}| |z_{2}| \sin(\arg(z_{1})) \sin(\arg(z_{2})).$$ By trig identities, this is equal to $$z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos(\arg(z_{1}) - \arg(z_{2})).$$ Since $|z_{1}| \neq 0$ and $|z_{2}| \neq 0$ (for, otherwise, $z_{1}$ and $z_{2}$ are trivially perpendicular), we must have $\cos(\arg(z_{1}) - \arg(z_{2})) = 0$, which can only happen if $\arg(z_{1}) - \arg(z_{2}) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.
For the converse, just work it backwards. We have $\arg(z_{1}) - \arg(z_{2})) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$, so $z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos( \arg(z_{1}) - \arg(z_{2})) = 0$.
So $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$.
If you prefer a geometrical vision, lock at the figure.
The area of the parallelogram $OMPN$ is obviously the same as the area of the pink parallelogram (same basis $OM$ and same height because the opposite side are parallel). But the area of the pink parallelogram is also given by:$ \quad Area=|a_1c|$ and we can find $c$ as the intersect at the origin of the line from $N,P$.
Whith a bit of elemental analitic geometry you can find that $$ c=\frac{a_1b_2-a_2b_1}{a_1} $$
and this menas that:
$$ Area= |a_1b_2-a_2b_1| $$
that is the absolute value of the derminat of the matrix that has as columns the two vectors $\overline {OM}$ and $\overline {ON}$ or the absolut value of the cross product of the same two vectors.
Best Answer
The length of the crossproduct equals the area of the parallelogram, spaned by its vectors.
$$A = \frac{1}{2} |v_1 \times v_2| $$ with $v_1 = (x_0 - x_1, y_0 - y_1, z_0 - z_1)^T$ and $v_1 = (x_0 - x_2, y_0 - y_2, z_0 - z_2)^T$
If you wonder what $| | $ means: it is the length of vector inside. In our case, the length of the vector of the crossproduct.
Now take a pen and a paper, write this down and derive the formula yourself