Question Number 53450 by ajfour last updated on 22/Jan/19
Commented by ajfour last updated on 22/Jan/19
$${Find}\:{radius}\:{of}\:{inscribed}\:{sphere}\:{in} \\ $$$${box}. \\ $$
Answered by ajfour last updated on 22/Jan/19
![let D be origin. x axis be DA, y axis be DC, z axis be DF. E(a,0,2p) ; B(a,2a,0); G(0,2a,p); F(0,0,3p). BG×BE = (−ai+pk)×(−2aj+2pk) ⇒ n^ = 2a^2 k+2apj+2api eq. of ceiling plane is (r^ −3pk).n^ =0 ⇒ (r^ −3pk^ ).(pi^ +pj^ +ak^ )=0 ⇒ r^ .(pi^ +pj^ +ak^ )=3ap center of sphere: G(r,r,r) ⇒ [r(i+j+k)+((rn^ )/(∣n^ ∣))].(pi+pj+ak)=3ap ⇒ r = ((3ap)/(2p+a+(√(2p^2 +a^2 )))) .](Q53458.png)
$${let}\:{D}\:{be}\:{origin}.\:{x}\:{axis}\:{be}\:{DA}, \\ $$$${y}\:{axis}\:{be}\:{DC},\:{z}\:{axis}\:{be}\:{DF}. \\ $$$${E}\left({a},\mathrm{0},\mathrm{2}{p}\right)\:;\:{B}\left({a},\mathrm{2}{a},\mathrm{0}\right);\:{G}\left(\mathrm{0},\mathrm{2}{a},{p}\right); \\ $$$${F}\left(\mathrm{0},\mathrm{0},\mathrm{3}{p}\right). \\ $$$${BG}×{BE}\:=\:\left(−{ai}+{pk}\right)×\left(−\mathrm{2}{aj}+\mathrm{2}{pk}\right) \\ $$$$\:\Rightarrow\:\:\bar {{n}}\:=\:\mathrm{2}{a}^{\mathrm{2}} {k}+\mathrm{2}{apj}+\mathrm{2}{api} \\ $$$${eq}.\:{of}\:{ceiling}\:{plane}\:{is} \\ $$$$\:\:\:\left(\bar {{r}}−\mathrm{3}{pk}\right).\bar {{n}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\bar {{r}}−\mathrm{3}{p}\hat {{k}}\right).\left({p}\hat {{i}}+{p}\hat {{j}}+{a}\hat {{k}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\bar {{r}}.\left({p}\hat {{i}}+{p}\hat {{j}}+{a}\hat {{k}}\right)=\mathrm{3}{ap} \\ $$$${center}\:{of}\:{sphere}:\:{G}\left({r},{r},{r}\right) \\ $$$$\Rightarrow\:\left[{r}\left({i}+{j}+{k}\right)+\frac{{r}\bar {{n}}}{\mid\bar {{n}}\mid}\right].\left({pi}+{pj}+{ak}\right)=\mathrm{3}{ap} \\ $$$$\Rightarrow\:{r}\:=\:\frac{\mathrm{3}{ap}}{\mathrm{2}{p}+{a}+\sqrt{\mathrm{2}{p}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\:. \\ $$
Commented by mr W last updated on 22/Jan/19
$${best}\:{way}\:{in}\:{this}\:{case}! \\ $$