Question Number 189807 by normans last updated on 22/Mar/23
Answered by a.lgnaoui last updated on 22/Mar/23
$$\: \\ $$$${Angle}\:{entre}\:{verticale}\:{et}\:{p}\:{q}\left(\alpha={arc}\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{6}}\right) \\ $$$${Angle}\:{entre}\:{Eq}\:{et}\:{qr}\:{est}\:\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\pi}{\mathrm{6}}+{X}+\frac{\pi}{\mathrm{4}}=\pi\Rightarrow\:\:{X}=\frac{\mathrm{7}\pi}{\mathrm{12}} \\ $$
Commented by a.lgnaoui last updated on 22/Mar/23
Answered by cortano12 last updated on 22/Mar/23
$$\mathrm{R}\left(\mathrm{2},\mathrm{1},\mathrm{0}\right)\:,\mathrm{Q}\left(\mathrm{2},\mathrm{0},\mathrm{1}\right),\mathrm{P}\left(\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$$\:\begin{cases}{\overset{\rightarrow} {\mathrm{QR}}\:=\:\left(\mathrm{0},\mathrm{1},−\mathrm{1}\right)}\\{\overset{\rightarrow} {\mathrm{QP}}\:=\left(−\mathrm{2},\mathrm{1},\mathrm{1}\right)}\end{cases} \\ $$$$\:\Rightarrow\mathrm{cos}\:\mathrm{X}\:=\:\frac{\overset{\rightarrow} {\mathrm{QR}}\:.\overset{\rightarrow} {\mathrm{QP}}}{\mid\overset{\rightarrow} {\mathrm{QR}}\mid\:.\mid\overset{\rightarrow} {\mathrm{QP}}\mid}\: \\ $$$$\Rightarrow\:\mathrm{cos}\:\mathrm{X}=\frac{\mathrm{0}+\mathrm{1}−\mathrm{1}}{\:\sqrt{\mathrm{2}}.\sqrt{\mathrm{6}}}=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{X}\:=\:\mathrm{90}° \\ $$
Answered by mr W last updated on 22/Mar/23
$${say}\:{edge}\:{length}\:{of}\:{cube}\:{is}\:\mathrm{2}. \\ $$$${qr}=\sqrt{\mathrm{2}} \\ $$$${qp}=\sqrt{\mathrm{6}} \\ $$$${qr}^{\mathrm{2}} +{qp}^{\mathrm{2}} =\mathrm{2}+\mathrm{6}=\mathrm{8} \\ $$$${pr}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${pr}^{\mathrm{2}} =\mathrm{8} \\ $$$${pr}^{\mathrm{2}} ={qr}^{\mathrm{2}} +{qp}^{\mathrm{2}} \\ $$$$\Rightarrow\Delta{pqr}\:{is}\:{right}\:{angled}.\:\Rightarrow{x}=\mathrm{90}° \\ $$