Find the area of the plane y=3x enclosed within the cube formed by the planes x=y=z= [itex]\pm[/itex] 5
The Attempt at a Solution
Using the equation y=3x, I found two points (x,y)=(5,1.66) and (x,y)=(-5,-1.66), then by plugging these points in the distance formula, I got the magnitude of one dimension equal to 10.54. I feel that the size of the other dimension must also be 10.54, so the area should be 111.09. But the answer given in my book is 105.4. I've been visualizing and thinking for hours on end about this, but I can't pin down what mistake I'm making.
It's such a simple problem, and I'm really very disappointed at my inability to think clearly. :/
Please help me.