asked 2014-07-31 17:38:09 -0600
Let D denote the solid pyramid with vertices located at (4,0,0), (0,1,0), (0,0,2), and the origin. Set up $\iiint_{D} f(x,y,z)dV$. So when we were talking about this in class we drew the graph first and then set up the equation ax+by+cz=d and then d was substituted for 4 because it was the least common multiple of the points. My question is how were the other numbers found in the equation which made it x+4y+2z=4 ?
answered 2014-07-31 19:11:19 -0600
Ok, you take each point one at a time and get your values for a,b,c.
Starting with $ax + by + cz = 4$ and the point $(4,0,0)$, you see that with this point $x=4, y=0$ and $ z=0$.
So we get $a4 +b0 + c0 = 4$
This leaves $4a=4$, so $a=1$.
Now with the point $(0,1,0) x=0, y=1$ and $z=0 $. So we have $by=4$, so $b=4$.
With $(0,0,2)$, $x=0, y=0$ and $z=2$ leaving $cz=4$, so $c=2$.
Now with values for a,b,c, we have the equation for the plane, $x + 4y +2z = 4$.
I never would have thought of doing this and thought it was pretty neat!
answered 2014-07-31 22:59:26 -0600
To add to this, when in doubt, you can always go back to the books and just do good old cross product to get the equation of plane. Since you have the three points, you can easily make two vectors where you can cross the two to find the perpendicular vector for the plane. The vectors I used were $ \langle 4, -1, 0 \rangle $ and $ \langle 0, -1, 2 \rangle $
When I crossed these two vectors, I came up with the same $x + 4y + 2z = 0$ equation that he used in class (multiplied by -2 or something)
Asked: 2014-07-31 17:38:09 -0600
Seen: 25 times
Last updated: Jul 31 '14
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