# Thread: Using vectors to find the volume of a pyramid

1. ## Using vectors to find the volume of a pyramid

Hi,

So here's my problem :

The equation of a plane is 3x+2y-6z=12

Which meets the x,y,and z-axis at points A, B, and C, which I calculated to be (4,0,0), (0,6,0), and (0,0,-2) respectively (I think I'm correct).

Now I have to find the volume of the pyramid OABC. I'm guessing it's a triangle based pyramid with the peak at O (origin). I already found the area of the base to be 14 using the fact that A=1/2*magnitude(AC x AB).

Now I'm stuck on how to find the height of the pyramid?

EDIT: Or actually what just came to mind is the fact that the plane is 12 units from the origin can I use that fact as the height? (might have just solved it /facepalm/)

Any help will be appreciated. Thanks.

2. Does your syllabus cover the following formula?

$\displaystyle V=\dfrac{1}{6}\mod \begin{vmatrix} {x_1}&{y_1}&{z_1}&{1}\\{x_2}&{y_2}&{z_2}&{1}\\{x_3 }&{y_3}&{z_3}&{1}\\{x_4}&{y_4}&{z_4}&{1}\end{vmatr ix}$

3. I don't think so, I mean I've never seen it in our data booklet (IB) or my teachers notes.

4. Then, use that the height of the pyramid is $\displaystyle 2$ .

5. No, Fernando, not with FinnSkies's choice of base. If you take the base to be the triangle with vertices (4, 0, 0), 0,6,0), and (0,0,-2) then the height is the perpendicular distance from that plane to (0, 0, 0). That plane can be written as $\displaystyle \frac{x}{4}+ \frac{y}{6}- \frac{z}{2}= 1$ and so $\displaystyle \frac{1}{4}\vec{i}+ \frac{1}{6}\vec{j}- \frac{1}{2}\vec{k}$ is normal to the plane. The line given by x= (1/4)t, y= (1/6)t, z= -(1/2)t is parallel to that vector and passes through (0, 0, 0). It intersects the plane when (1/16)t+ (1/36)t+ (1/4)t= \frac{49}{144}t= 1. t= 144/49 so that the point of intersection is (36/49, 4/49, -72/49). The height of the pyramid is the distance from (0, 0, 0) to that point, $\displaystyle \frac{4}{7}\sqrt{406}$.

Of course, it is simpler to use one of the other planes as base. If we take the triangle with vertices at (0, 0, 0), (4, 0, 0), and (0, 6, 0) as base, then, because the third vertex, (0, 0, -2), is on the z-axis, perpendicular to that plane, the height is just 2.

6. Originally Posted by HallsofIvy
No, Fernando, not with FinnSkies's choice of base.

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