How to Find the Volume of a Tetrahedron?

[Saitama]

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!

 

[I like Serena]

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?

 

[Saitama]

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:
$$V=\frac{1}{6}\left|\left[\vec{c} \vec{b} \vec{d}\right]\right|=\frac{3}{6}=\frac{1}{2}$$
which is the correct answer.

Thanks a lot ILS! :)

 

[MarkFL]

...which is the correct answer.

Thanks a lot ILS! :)

Thank you for taking the time to show how the help given allowed you to resolve the question! (Yes)

 

[CellCoree]

Dear [Name],

Thank you for sharing your attempt at finding the volume of a tetrahedron with the given information. It seems like you have made good progress so far by setting up the equations and using the given values to find the angle and distance between the two skew lines.

To find the volume of the tetrahedron, we can use the formula V = (1/3) * base area * height. In this case, the base of the tetrahedron is the triangle formed by points A, B, and C, and the height is the perpendicular distance between the base and point D.

To find the base area, we can use the cross product of vectors AB and AC, which will give us the area of the parallelogram formed by these two vectors. Then, we can divide this by 2 to get the area of the triangle.

Next, we need to find the height of the tetrahedron. To do this, we can use the formula for the distance between a point and a line, which is given by d = |(P - A) * n| / |n|, where P is the point, A is a point on the line, and n is the direction vector of the line. In our case, P is point D, A is any point on the line AB, and n is the cross product of vectors AB and CD.

Once we have the base area and height, we can plug them into the formula for the volume of a tetrahedron to get our final answer.

I hope this helps guide you in the right direction. If you have any further questions, please do not hesitate to ask.

Best,