Question Number 191935 by Rupesh123 last updated on 04/May/23
Answered by som(math1967) last updated on 04/May/23
$${Direction}\:{cosine}\:{of}\:{L} \\ $$$${cos}\alpha,{cos}\beta,{cos}\gamma \\ $$$$\:\therefore{cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}^{\mathrm{2}} \gamma=\mathrm{1} \\ $$$$\mathrm{1}−{sin}^{\mathrm{2}} \alpha+\mathrm{1}−{sin}^{\mathrm{2}} \beta+\mathrm{1}−{sin}^{\mathrm{2}} \gamma=\mathrm{1} \\ $$$$\therefore{sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \gamma=\mathrm{3}−\mathrm{1}=\mathrm{2} \\ $$
Answered by mehdee42 last updated on 04/May/23
$${cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}^{\mathrm{2}} \gamma=\mathrm{1}\Rightarrow \\ $$$${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \gamma=\mathrm{2} \\ $$$${tip}\:\because\:{cos}\alpha=\frac{{a}}{\mid{v}\mid}\:\:\&\:{cos}\beta=\frac{{b}}{\mid{v}\mid}\:\&\:{cos}\gamma=\frac{{c}}{\mid{v}\mid}\:;\:{v}=\left({a},{b},{c}\right) \\ $$