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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

**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!