If $\alpha$, $\beta$, $\gamma$, the angles made by a vector with $x, y, z$ axes, are in the ratio 1:2:3 then their values are?

Question

This is a question asked in a book:

If $\alpha$, $\beta$, $\gamma$ are angles made by a vector $x, y, z$ axes respectively, and $\alpha$:$\beta$:$\gamma$ = 1:2:3 then values of $\alpha$, $\beta$, $\gamma$ are?

I tried the obvious by taking $\alpha$ = k; $\beta$=2k; $\gamma$ = 3k; and substituting in $cos^2\alpha$+$cos^2\beta$+$cos^2\gamma$ = 1

but its complicating things. I'm must be missing something simple and obvious but I can't figure out what.

Answer 1

If $\cos^2 x +\cos^2(2x)+\cos^2(3x)=1 $, then, since $\cos^2(z) =(\cos(2z)+1)/2 $, $1 =(\cos(2x)+\cos(4x)+\cos(6x)+3)/2 $ or $\cos(2x)+\cos(4x)+\cos(6x) = -1 $.

Letting $y = 2x$,

$\begin{array}\\ -1 &=\cos(y)+\cos(2y)+\cos(3y)\\ &=\cos(y)+2\cos^2y-1+4\cos^3(y)-3\cos(y)\\ &=4\cos^3(y)+2\cos^2y-2\cos(y)-1\\ \end{array} $

or $4\cos^3(y)+2\cos^2y-2\cos(y) =0 $.

Letting $z = \cos y$,

$\begin{array}\\ 0 &=2z^3+2z^2-z\\ &=z(2z^2+z-1)\\ &=z(z+1)(2z-1) \qquad\text{just fixed an error here}\\ \end{array} $

so $z = 0, -1, \frac12 $.

Therefore $y=\pi/2, \pi, \pi/3$ so $x = y/2 =\pi/4, \pi/2, \pi/6 $.

Answer 2

Taking vector $(\cos(30°),\sin(30°),0)=(\sqrt{3}/2,1/2,0)$, you get the angles $30°, 60°, 90°.$

Answer 3

It seems to me that we cannot avoid to solve the equation $$ \cos^2 x +\cos^2(2x)+\cos^2(3x)=1 $$ For $x$ between $0$ and $\pi/2$ Wolfram Alpha gives the solution $x=\frac{\pi}{4}$ (that is an obvious solution).