# identity

• Feb 10th 2008, 02:08 AM
heathrowjohnny
identity
How do you prove the following: Given the direction cosines of two lines $\displaystyle \cos \alpha, \ \cos \beta, \ \cos \gamma$ and $\displaystyle \cos \alpha ', \cos \beta ', \cos \gamma ',$, then the cosine of the angle $\displaystyle \theta$ between the two lines is $\displaystyle \cos \theta = \cos \alpha \cos \alpha ' + \cos \beta \cos \beta ' + \cos \gamma \cos \gamma '$?
• Feb 10th 2008, 04:31 AM
Soroban
Hello, heathrowjohnny!

Quote:

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

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

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

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

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

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