# Vector Angle

• Oct 16th 2011, 05:23 AM
BabyMilo
Vector Angle
As below
• Oct 16th 2011, 09:06 AM
Soroban
Re: Vector Angle
Hello, BabyMilo!

Quote:

$\displaystyle \text{The vector }\overline{OP}\text{ makes angle of }60^o\text{ with the positive }x\text{-axis}$
. . $\displaystyle \text{and }45^o\text{ with the positive }y\text{-axis.}$
$\displaystyle \text{Find the possibles angles the vector can make with the positve }z\text{-axis.}$
. . . . . . . . . . $\displaystyle \text{Theorem}$

$\displaystyle \text{If }\alpha,\,\beta,\,\gamma\text{ are the direction angles of a vector,}$
. . . . then: .$\displaystyle \cos^2\!\alpha + \cos^2\!\beta + \cos^2\!\gamma \;=\;1$

We have: .$\displaystyle \left(\cos 60^o)^2 + (\cos45^o)^2 + \cos^2\gamma \;=\;1$

. w . . . . . . . . . . . $\displaystyle \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{\sqrt{2}}\right)^2 + \cos^2\gamma \;=\;1$

. a . . . . . . . . . . . . . . . . . $\displaystyle \tfrac{1}{4} + \tfrac{1}{2} + \cos^2\gamma \;=\;1$

. m . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \cos^2\gamma \;=\;\tfrac{1}{4}$

. . n . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \cos\gamma \;=\; \pm\tfrac{1}{2}$

. . . . . . . . . . . . . . . . . .Therefore: n . $\displaystyle \gamma \;=\;60^o,\,120^o$