The volume of a tetrahedron is (1/3)*(area of base)*(height).

Use this result to prove that the volume of a tetrahedron whose edges are the vectors u, v, w is (1/6) lu dot (v x w)l.

note: dot = dot product.

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- Jan 16th 2007, 09:25 AMJenny20a proof
The volume of a tetrahedron is (1/3)*(area of base)*(height).

Use this result to prove that the volume of a tetrahedron whose edges are the vectors u, v, w is (1/6) lu dot (v x w)l.

note: dot = dot product. - Jan 16th 2007, 10:14 AMCaptainBlack
is a vector normal to the plane containing and and whose magnitude is equal to the area of the parallelogram defined by them, and so twice the area of the triangle defined by them.

Now the absolute value of is equal to the product of the projection of onto , which is the height of the tetrahedron times twice the area of the base (upto the sign).

Thus , is equal to the height of the tetrahedron times twice the area of the base, which is six times the volume of the tetrahedron, which is what was to be proven.

RonL - Jan 16th 2007, 12:03 PMSoroban
Hello, Jenny!

My proof is the same as Captain Black's with some small differences.

Quote:

The volume of a tetrahedron is: .

Use this result to prove that the volume of a tetrahedron

whose edges are the vectors is: .

Let the plane of and be the base of the tetrahedron.

Then: .

The height of the tetradedron is the magnitude of the projection of onto

. . The projection of onto is given by: .

We have: .

. . Then: .

Substitute: .

. . . . . . . . .

Therefore: .

- Jan 16th 2007, 01:31 PMJenny20
Hi Soroban,

With Captainblack's help , i got the same proof as you. :)