Question

$if \cos( \alpha ). \: \cos( \beta ) . \: \cos( \gamma ) \: are \: the \: direction \: cosine \: of \: a \: line \: then \: find \: the \: value \: of \: \sin^{2} \alpha + \sin^{2} \beta + \sin^{2} \gamma$
Please answer ​

Answer 1

AnswEr :

Given.

$\sf \: \cos( \alpha ) , \cos( \beta ) \: and \: \cos( \gamma ) \: are \: directional \: cosine \: of \: a \: straight \: line.$

Let the line = ai + bj + ck and r be the magnitude of the line.

Now,

$\sf \: r = \sqrt{a {}^{2} + {b}^{2} + {c}^{2} }$

Also,

$\sf \: \cos( \alpha ) = \dfrac{a}{r} - - - - - (1) \\ \sf \cos( \beta ) = \dfrac{b}{r} - - - - - (2) \\ \: \ \sf\cos( \gamma ) = \dfrac{c}{r} - - - - - - (3)$

Squaring the above system of equations and adding them,

$\implies \sf \: \cos {}^{2} ( \alpha ) + \cos {}^{2} ( \beta ) + \cos {}^{2} ( \gamma ) \\ \\ \implies \sf \: \dfrac{ {a}^{2} }{ {r}^{2} } + \dfrac{b {}^{2} }{r {}^{2} } + \dfrac{ {c}^{2} }{r {}^{2} } \\ \\ \implies \sf \: \dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {r}^{2} } \\ \\ \implies \sf \: \dfrac{r {}^{2} }{ {r}^{2} } \\ \\ \implies \: 1$

Thus,

$\longrightarrow \sf \: \cos {}^{2} ( \alpha ) + \cos {}^{2} ( \beta ) + \cos {}^{2} ( \gamma ) = 1 \\ \\ \longrightarrow \sf \: 1 - \sin {}^{2} ( \alpha ) + 1 - \sin {}^{2} ( \beta ) + 1 - \sin {}^{2} ( \gamma ) = 1 \\ \\ \longrightarrow \sf \: \sin {}^{2} ( \alpha ) + \sin {}^{2} ( \beta ) + \sin {}^{2} ( \gamma ) = 3 - 2 \\ \\ \longrightarrow \boxed{ \boxed{\sf \: \sin {}^{2} ( \alpha ) + \sin {}^{2} ( \beta ) + \sin {}^{2} ( \gamma ) = 1}}$

Answer 2

Hope u understood...This is the correct answerr