The following is the question in my textbook:-
If a straight line makes angle $\alpha$, $\beta$, $\gamma$ with the $x, y, z$ axes respectively, then show that $\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma} = 2 $?
Here is what I have done:-
Since $\alpha$, $\beta$, $\gamma$ are the angles made by the line with the axes so $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are the direction cosines of the line
now
$$\sin^2\theta + \cos^2\theta = 1$$
$$\sin^2\theta = 1 – \cos^2\theta$$
so
$$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 1 – \cos^2\alpha + 1 – \cos^2\beta + 1 – \cos^2\gamma$$$$ =3 – (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$$
now since $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are the direction cosines of the line so $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
so
$$3 – \cos^2\alpha + \cos^2\beta + \cos^2\gamma= 3 – 1$$
$$=2- answer$$
My question was that have I done it correctly and if not what is the correct way of doing it.
Best Answer
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ It's completely general when we choose that the line contains the origin of coordinates. In spherical coordinates $\ds{\pars{~0 \leq \theta < \pi\,,\quad 0 \leq \phi < 2\pi~}}$: \begin{align} &\color{#66f}{\large\sin^{2}\pars{\alpha} + \sin^{2}\pars{\beta} + \sin^{2}\pars{\gamma}} \\[3mm]&=\overbrace{\bracks{1 - \sin^{2}\pars{\theta}\cos^{2}\pars{\phi}}} ^{\ds{\sin^{2}\pars{\alpha}}}\ +\ \overbrace{\bracks{1 - \sin^{2}\pars{\theta}\sin^{2}\pars{\phi}}} ^{\ds{\sin^{2}\pars{\beta}}}\ +\ \overbrace{\bracks{1 - \cos^{2}\pars{\theta}}}^{\ds{\sin^{2}\pars{\gamma}}} \\[3mm]&=3 - \sin^{2}\pars{\theta} \bracks{\cos^{2}\pars{\phi} + \sin^{2}\pars{\phi}} - \cos^{2}\pars{\theta} \\[3mm]&=3 - \sin^{2}\pars{\theta} - \cos^{2}\pars{\theta} =3 - 1 = \color{#66f}{\Large 2} \end{align}