Question Number 10392 by ketto last updated on 06/Feb/17
$${find}\:{the}\:{direction}\:{cosines}\: \\ $$$${and}\:{its}\:{angles}\:{on} \\ $$$$\mathrm{2}{i}\:−\:\mathrm{3}{j}\: \\ $$
Answered by mrW1 last updated on 07/Feb/17
$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$\mathrm{cos}\:\beta=\frac{−\mathrm{3}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }}=−\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}} \\ $$$$\alpha=\mathrm{56}.\mathrm{3}° \\ $$$$\beta=\mathrm{146}.\mathrm{3}° \\ $$