The equation of a plane in space is x + 3y + 2z = 6. 1.) The three planes of the axes and the plane you have sketched create a triangular pyramid. Find the volume of this pyramid, showing the formula you are using and all calculations below. For my intercepts I got values of, (6, 2, 3), with x = 6, y = 2, and z = 3. I plotted these intercepts on the graph. How would I do part B of this problem (finding the volume)?

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The equation of a plane in space is x + 3y + 2z = 6. 1.) The three planes of the axes and the plane you have sketched create a triangular pyramid. Find the volume of this pyramid, showing the formula you are using and all calculations below. For my intercepts I got values of, (6, 2, 3), with x = 6, y = 2, and z = 3. I plotted these intercepts on the graph. How would I do part B of this problem (finding the volume)?

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are you sure you are on track? maybe you are but thought i would ask write the plane \( x + 3y + 2z = 6 \) as \(z = 3 - \frac{x}{2} - \frac{3y}{2}\) gives |dw:1439328868573:dw|
OK i now see that you have just presented your intercepts differently/ unconventionally [as if they were one] the next step is to integrate \(\large \int \int z(x,y) \ dA = \int \int 3 - \frac{x}{2} - \frac{3y}{2} \ dx \ dy\) which is either: \(\large \int_{y=0}^{2} \int_{x = 0}^{6 - 3y} 3 - \frac{x}{2} - \frac{3y}{2} \ dx \ dy\) or \(\large \int_{x=0}^{6} \int_{y = 0}^{2 -\frac{x}{3}} 3 - \frac{x}{2} - \frac{3y}{2} \ dy \ dx\) i get 6.
@Irishboy123 If I'm going to be quite honest with you... I have no idea what most of that means. I'm taking a geometry course as a sophomore in high school. Could you explain that set of equations to me in a simpler way?

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\[volume=\frac{ 1 }{ 3 }\left( area ~of~base \right)\times height\]
Would this be the correct equation, then? \[\frac{ 1 }{ 3 } (6) (3) = 6\]
i think so.
Assuming that my calculation of 6 equaling the base area is correct.

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