Vector Angle

**BabyMilo** · October 16th 2011, 05:23 AM

As below

**Soroban** · October 16th 2011, 09:06 AM

Hello, BabyMilo!

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

. . . . . . . . . . $\text{Theorem}$

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

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

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

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

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

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

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

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