1. ## a 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.

2. Originally Posted by Jenny20
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.
$\bold{v} \wedge \bold{w}$ is a vector normal to the plane containing $\bold{v}$ and $\bold{w}$ 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 $\bold{u} \vee (\bold{v} \wedge \bold{w})$ is equal to the product of the projection of $\bold{u}$ onto $\bold{v} \wedge \bold{w}$, which is the height of the tetrahedron times twice the area of the base (upto the sign).

Thus $|\bold{v} \wedge \bold{w}|$, 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

3. Hello, Jenny!

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

The volume of a tetrahedron is: . $\frac{1}{3}(\text{area of base})(\text{height})$

Use this result to prove that the volume of a tetrahedron
whose edges are the vectors $\vec{u},\:\vec{v},\:\vec{w}$ is: . $\frac{1}{6}\bigg|u \cdot (v \times w)\bigg|$

Let the plane of $\vec{v}$ and $\vec{w}$ be the base of the tetrahedron.

Then: . $(\text{area of base}) \:=\:\frac{1}{2}\left|\vec{v} \times \vec{w}\right|$

The height of the tetradedron is the magnitude of the projection of $\vec{u}$ onto $\vec{v} \times \vec{w}$

. . The projection of $\vec{a}$ onto $\vec{b}$ is given by: . $\text{proj}_{\vec{b}}\vec{a} \;=\;\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2} \right)\vec{b}$

We have: . $\vec{h} \:=\:\left(\frac{\vec{u}\cdot(\vec{v} \times \vec{w})}{|\vec{v} \times \vec{w}|^2}\right)(\vec{v} \times \vec{w})$

. . Then: . $h \;=\;\left|\frac{\vec{u}\cdot(\vec{w}\times\vec{w} )}{|\vec{v}\times\vec{w}|^2}\,(\vec{v}\times\vec{w })\right| \;=\;\frac{|\vec{u}\cdot(\vec{v}\times\vec{w})|}{| \vec{v}\times\vec{w})|^2}\,|\vec{v}\times\vec{w}|$ $= \;\frac{|\vec{u}\cdot(\vec{v}\times\vec{w})|}{|\ve c{v}\times\vec{w}|}$

Substitute: . $V \;=\;\frac{1}{3}\underbrace{(\text{area of base})}\underbrace{(\text{height})}$

. . . . . . . . . $V \;=\;\frac{1}{3}\left[\frac{1}{2}|\vec{v}\times\vec{w}|\right] \,\left[\frac{|\vec{u}\cdot(\vec{v}\times\vec{w})|}{|\vec{ v}\times\vec{w}|}\right]$

Therefore: . $V \;=\;\frac{1}{6}\bigg|\vec{u}\cdot(\vec{v}\times\v ec{w})\bigg|$

4. Hi Soroban,

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

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### volume of tetrahedron vectors proof

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