# Volume of tetrahedron

#### Pranav

##### Well-known member
Problem:
Suppose in a tetrahedron ABCD, AB=1; CD=$\sqrt{3}$; the distance and the angle between the skew lines AB and CD are 2 and $\pi/3$ respectively. Find the volume of tetrahedron.

Attempt:
Let the points A,B,C and D be represented by the vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ respectively. Then, as per the question, I have:
$$\left|\vec{b}-\vec{a}\right|=1$$
$$\left|\vec{d}-\vec{c}\right|=\sqrt{3}$$
The line AB can be represented as $\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$ and the line CD can be represented by $\vec{r}=\vec{c}+\mu (\vec{d}-\vec{c})$ where $\lambda$ and $\mu$ are scalars. The angle ($\theta$) between the two lines is given by:
$$\cos\theta=\frac{(\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})}{\left|\vec{b}-\vec{a}\right| \left|\vec{d}-\vec{c}\right|}$$
$$\Rightarrow \frac{1}{2}=\frac{(\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})}{\sqrt{3}}$$
$$\Rightarrow (\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})=\frac{\sqrt{3}}{2}$$
The distance between the two lines is 2 so I have the following relation:
$$\left|\frac{(\vec{a}-\vec{c})\cdot ((\vec{b}-\vec{a})\times (\vec{d}-\vec{c})}{\left|(\vec{b}-\vec{a})\times(\vec{d}-\vec{c})\right|}\right|=2$$
$$\Rightarrow \left|\left[\vec{a}-\vec{c}\,\,\,\, \vec{b}-\vec{a}\,\,\,\, \vec{d}-\vec{c}\right]\right|=3$$

I am clueless about the next step.

Any help is appreciated. Thanks!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Problem:
Suppose in a tetrahedron ABCD, AB=1; CD=$\sqrt{3}$; the distance and the angle between the skew lines AB and CD are 2 and $\pi/3$ respectively. Find the volume of tetrahedron.

Attempt:
Let the points A,B,C and D be represented by the vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ respectively. Then, as per the question, I have:
$$\left|\vec{b}-\vec{a}\right|=1$$
$$\left|\vec{d}-\vec{c}\right|=\sqrt{3}$$
The line AB can be represented as $\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})$ and the line CD can be represented by $\vec{r}=\vec{c}+\mu (\vec{d}-\vec{c})$ where $\lambda$ and $\mu$ are scalars. The angle ($\theta$) between the two lines is given by:
$$\cos\theta=\frac{(\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})}{\left|\vec{b}-\vec{a}\right| \left|\vec{d}-\vec{c}\right|}$$
$$\Rightarrow \frac{1}{2}=\frac{(\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})}{\sqrt{3}}$$
$$\Rightarrow (\vec{b}-\vec{a})\cdot (\vec{d}-\vec{c})=\frac{\sqrt{3}}{2}$$
The distance between the two lines is 2 so I have the following relation:
$$\left|\frac{(\vec{a}-\vec{c})\cdot ((\vec{b}-\vec{a})\times (\vec{d}-\vec{c})}{\left|(\vec{b}-\vec{a})\times(\vec{d}-\vec{c})\right|}\right|=2$$
$$\Rightarrow \left|\left[\vec{a}-\vec{c}\,\,\,\, \vec{b}-\vec{a}\,\,\,\, \vec{d}-\vec{c}\right]\right|=3$$

I am clueless about the next step.

Any help is appreciated. Thanks!
Hey Pranav!

Shall we pick $\vec a = \vec 0$?
That makes those formulas a bit easier.

Do you also have a formula for the volume of a tetrahedron in terms of those vectors?

#### Pranav

##### Well-known member
Hey Pranav!

Shall we pick $\vec a = \vec 0$?
That makes those formulas a bit easier.
Completely agreed and I seem to have reached the answer too because of that.

Let $\vec{a}=0$. So the formulas I posted above becomes:
$$\left|\vec{b}\right|=1$$
and
$$\left|\left[\vec{c} \vec{b} \vec{d}-\vec{c}\right]\right|=3$$
The above scalar triple product is same as:
$$\left|\left[\vec{c} \vec{b} \vec{d}\right]\right|=3$$
Do you also have a formula for the volume of a tetrahedron in terms of those vectors?
Yes. The volume of tetrahedron in the given case is:
$$V=\frac{1}{6}\left|\left[\vec{c}-\vec{a} \vec{b}-\vec{a} \vec{d}-\vec{a}\right]\right|$$
But since we let $\vec{a}=0$, the above formula becomes:
$$V=\frac{1}{6}\left|\left[\vec{c} \vec{b} \vec{d}\right]\right|=\frac{3}{6}=\frac{1}{2}$$