Area of $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Question

Could you tell me how to calculate the area of part of the plane: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, $a, b, c >0$ where all coordinates of a point are positive?

Answer 1

Hint: solve for when $2$ of $x,y,$ and $z$ are $0$. This will give you the $3$ coordinates in $\mathbb{R}^3$ which are the vertices of the triangle you are looking for.

Answer 2

It's a triangle. Find the vectors along two sides and use the relation which connects vectors and the area of the triangle which they border. In particular, if $v,w$ are the vectors along the side then $\frac{1}{2}|v \times w|$ is the area. So, find those vectors. Hint: $x=y=0$ for $v$ and $y=z=0$ for $w$.