Find the area of a $S$, the part of the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ in the first octant where $x,y,z>0$. Also the parameters $a,b,c>0$. The final answer should be: $$\sqrt{\frac{a^2b^2}{4}+\frac{a^2c^2}{4}+\frac{b^2c^2}{4}}$$
I think there're several ways to do it, but I think this could be solved by doing vector product. We know the norm of vector product gives us parallelogram so to get the area of $S$, (triangle) we need to divide the final answer by $2$.
The problem I'm running into is determining the two vectors.
First, I tried getting the points $a,b,c$ (this is bit confusing as these are also the parameters in the plane equation but they appear in the drawing). So I think it should be: $$ b=\left(0,\frac{y}{b},0\right)\\ a=\left(\frac{x}{a},0,0\right)\\ c=\left(0,0,\frac{z}{c}\right) $$ Which would produce vectors: $$ AB=\left\langle -\frac{x}{a},\frac{y}{b},0\right\rangle\\ AC=\left\langle -\frac{x}{a},0,\frac{z}{c}\right\rangle $$ But the vector product doesn't give the desired solution: $$ AB\times AC=\left\langle \frac{yz}{bc},-\frac{xz}{ac}, \frac{yx}{ba}\right\rangle $$ What am I missing?
The vertices of your triangle are $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$. So, what you have to do is to compute the vector product $\bigl((0,b,0)-(a,0,0)\bigr)\times\bigl((0,0,c)-(a,0,0)\bigr)$, which is $-(bc,ac,ab)$. It's norm is $\sqrt{a^2b^2+b^2c^2+c^2a^2}$ and halving it gives the result that you're after.