Direction of Angles

Nov 2009
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0
Not sure what to do here:

The direction angles of a vector are all equal. Find the direction angles to the nearest degree.
 

Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, Neconine!

The direction angles of a vector are all equal.
Find the direction angles to the nearest degree.

The direction angles of a vector are \(\displaystyle \alpha,\:\beta,\:\gamma\), which are the angles
. . formed by the vector and the \(\displaystyle x.\:y,\) and \(\displaystyle z\) axes, respectively.

Fact: .\(\displaystyle \cos^2\!\alpha + \cos^2\!\beta + \cos^2\!\gamma \:=\:1\)


Since \(\displaystyle \alpha = \beta = \gamma\) we have: .\(\displaystyle \cos^2\!\alpha + \cos^2\!\alpha + \cos^2\!\alpha \;=\;1\)

. . \(\displaystyle 3\cos^2\!\alpha \:=\:1 \quad\Rightarrow\quad \cos^2\!\alpha \:=\:\frac{1}{3} \quad\Rightarrow\quad \cos\alpha \:=\:\frac{1}{\sqrt{3}} \)
. . \(\displaystyle \alpha \;=\;\cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \;=\;54.73561032^o \)


Therefore: .\(\displaystyle \alpha \;=\;\beta\;=\;\gamma \;\approx\;55^o\)