Normal and surface area of a plane

Question

Find the area of the portion of the plane $2x+3y+4z=28$ lying above rectangle $1\le\ x \le 3$, $2 \le\ y\le5$ in the xy-plane. I know this is solved by $||N|| = \sqrt{1+g_x+g_y}\ $ where $g(x,y)=(x,y,7-\frac{x}2\ -\frac{3y}4\ )$. But how come I cannot simply use $<2,3,4>$ for the normal in the formula $\int_ {1}^{3} \int_{2}^{5} ||N(x,y)|| \,dx\,dy$ ?

Answer 1

The equation of the plane is not unique. For example, you could multiply the entire equation by 10 and still have the same plane but the normal vector would be 10 times longer, yielding 10 times more surface area with the given formulation. Recall that for the formula for surface area we need to start with a surface of the form $z = f(x,y)$ and work from there. Rearranging the plane equation into that form, we obtain $N = \langle -1/2, -3/4, 1 \rangle$, so $\|N\| = \sqrt{29}/4$.