identity

**heathrowjohnny** · February 10th 2008, 02:08 AM

How do you prove the following: Given the direction cosines of two lines $\cos \alpha, \ \cos \beta, \ \cos \gamma$ and $\cos \alpha ', \cos \beta ', \cos \gamma ',$ , then the cosine of the angle $\theta$ between the two lines is $\cos \theta = \cos \alpha \cos \alpha ' + \cos \beta \cos \beta ' + \cos \gamma \cos \gamma '$ ?

**Soroban** · February 10th 2008, 04:31 AM

Hello, heathrowjohnny!

Given the direction cosines of two lines: . $[\cos\alpha,\cos\beta,\cos\gamma]$ and $[\cos\alpha',\cos\beta',\cos\gamma']$ ,
then the cosine of the angle $\theta$ between the two lines is:
. . $\cos\theta \;= \;\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma'$

Line $L_1$ has vector: . $\vec{u} \:=\:\langle\cos\alpha,\,\cos\beta,\,\cos\gamma\ra ngle$ . . . and: . $|\vec{u}| \:=\:1$

Line $L_2$ has vector: . $\vec{v} \:=\:\langle\cos\alpha',\,\cos\beta',\,\cos\gamma' \rangle$ . . . and: . $|\vec{v}| \:=\:1$

. . That is, a set of direction cosines always forms a unit vector.

The angle $\theta$ between two vectors $\vec{u}\text{ and }\vec{v}$ is given by: . $\cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}| }$
So we have: . $\cos\theta \;=\;\frac{\langle\cos\alpha,\cos\beta,\cos\gamma\ rangle\cdot\langle\cos\alpha',\beta',\gamma'\rangl e}{(1)(1)}$

Therefore: . $\cos\theta \;=\;\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma'$

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