Question-188845

Question Number 188845 by ajfour last updated on 07/Mar/23

Answered by a.lgnaoui last updated on 07/Mar/23

△OAB (recrangle en O) AB=a+((a(√2))/2) ⇒AH=(a/2)+((a(√2))/4) OA=OB =R sin (π/4)=((AH)/R)⇒R=((2AH)/( (√2))) R=1+(√2)

$$\bigtriangleup{OAB}\:\:\:\left({recrangle}\:{en}\:{O}\right) \\ $$$${AB}={a}+\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\Rightarrow{AH}=\frac{{a}}{\mathrm{2}}+\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${OA}={OB}\:={R} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{4}}=\frac{{AH}}{{R}}\Rightarrow{R}=\frac{\mathrm{2}{AH}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{R}}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$

Commented by a.lgnaoui last updated on 07/Mar/23

Commented by mr W last updated on 08/Mar/23

$${clearly}\:{wrong}! \\ $$$${a}\:{ball}\:{within}\:{the}\:{cube}\:{can}\:{not}\:{be}\:{larger} \\ $$$${than}\:{the}\:{cube}\:{itself}! \\ $$

Answered by mr W last updated on 08/Mar/23

Commented by mr W last updated on 08/Mar/23

A(r,r,r) P(1,0,0) PQ^(→) =(−1,1,1) PA^(→) =(r−1,r,r) (r−1,r,r)×(−1,1,1)=(0,2r−1,2r−1) r=((√(2(2r−1)^2 ))/( (√3)))=distance from A to PQ 5r^2 −8r+2=0 ⇒r=((4−(√6))/5)≈0.3101 < 0.5 (within cube)

$${A}\left({r},{r},{r}\right) \\ $$$${P}\left(\mathrm{1},\mathrm{0},\mathrm{0}\right) \\ $$$$\overset{\rightarrow} {{PQ}}=\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\overset{\rightarrow} {{PA}}=\left({r}−\mathrm{1},{r},{r}\right) \\ $$$$\left({r}−\mathrm{1},{r},{r}\right)×\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)=\left(\mathrm{0},\mathrm{2}{r}−\mathrm{1},\mathrm{2}{r}−\mathrm{1}\right) \\ $$$${r}=\frac{\sqrt{\mathrm{2}\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}={distance}\:{from}\:{A}\:{to}\:{PQ} \\ $$$$\mathrm{5}{r}^{\mathrm{2}} −\mathrm{8}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{4}−\sqrt{\mathrm{6}}}{\mathrm{5}}\approx\mathrm{0}.\mathrm{3101}\:<\:\mathrm{0}.\mathrm{5}\:\left({within}\:{cube}\right) \\ $$

Commented by ajfour last updated on 08/Mar/23

(r−1,r,r)×(−1,1,1)=(0,2r−1,2r−1) i dint understand above..but PA.λ^� =(√(PA^2 −r^2 )) (1−r+r+r)^2 =3{(r−1)^2 +2r^2 −r^2 } (1+r)^2 =3r^2 +3(1−r)^2 ⇒ 5r^2 −8r+2=0 r=((8±(√(64−40)))/(10))=((4±(√6))/5) here r=((4−(√6))/5)

$$\left({r}−\mathrm{1},{r},{r}\right)×\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)=\left(\mathrm{0},\mathrm{2}{r}−\mathrm{1},\mathrm{2}{r}−\mathrm{1}\right) \\ $$$${i}\:{dint}\:{understand}\:{above}..{but} \\ $$$${PA}.\hat {\lambda}=\sqrt{{PA}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\:\left(\mathrm{1}−{r}+{r}+{r}\right)^{\mathrm{2}} =\mathrm{3}\left\{\left({r}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} −{r}^{\mathrm{2}} \right\} \\ $$$$\left(\mathrm{1}+{r}\right)^{\mathrm{2}} =\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{5}{r}^{\mathrm{2}} −\mathrm{8}{r}+\mathrm{2}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{40}}}{\mathrm{10}}=\frac{\mathrm{4}\pm\sqrt{\mathrm{6}}}{\mathrm{5}} \\ $$$${here}\:\:\:{r}=\frac{\mathrm{4}−\sqrt{\mathrm{6}}}{\mathrm{5}} \\ $$

Commented by mr W last updated on 08/Mar/23

PA^(→) ×PQ^(→) =(r−1,r,r)×(−1,1,1)=(0,2r−1,2r−1)

$$\overset{\rightarrow} {{PA}}×\overset{\rightarrow} {{PQ}}=\left({r}−\mathrm{1},{r},{r}\right)×\left(−\mathrm{1},\mathrm{1},\mathrm{1}\right)=\left(\mathrm{0},\mathrm{2}{r}−\mathrm{1},\mathrm{2}{r}−\mathrm{1}\right) \\ $$

Commented by mr W last updated on 08/Mar/23