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Question-211058




Question Number 211058 by mr W last updated on 27/Aug/24
Commented by mr W last updated on 27/Aug/24
find the radius R of the largest sphere  which can be placed between the red  and blue planes in the first octant.  assume 0<p≤a, 0<q≤b, 0<r≤c.
$${find}\:{the}\:{radius}\:{R}\:{of}\:{the}\:{largest}\:{sphere} \\ $$$${which}\:{can}\:{be}\:{placed}\:{between}\:{the}\:{red} \\ $$$${and}\:{blue}\:{planes}\:{in}\:{the}\:{first}\:{octant}. \\ $$$${assume}\:\mathrm{0}<{p}\leqslant{a},\:\mathrm{0}<{q}\leqslant{b},\:\mathrm{0}<{r}\leqslant{c}. \\ $$
Answered by mr W last updated on 29/Aug/24
Commented by mr W last updated on 28/Aug/24
the largest sphere above and touching  the lower plane (x/p)+(y/q)+(z/r)=1:  say its radius is R_o , then its center  is at (R_o ,R_o ,R_o ).  we have  (((R_o /p)+(R_o /q)+(R_o /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=R_o   ⇒R_o =(1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))
$${the}\:{largest}\:{sphere}\:{above}\:{and}\:{touching} \\ $$$${the}\:{lower}\:{plane}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{p}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{q}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{r}}}=\mathrm{1}: \\ $$$${say}\:{its}\:{radius}\:{is}\:{R}_{{o}} ,\:{then}\:{its}\:{center} \\ $$$${is}\:{at}\:\left({R}_{{o}} ,{R}_{{o}} ,{R}_{{o}} \right). \\ $$$${we}\:{have} \\ $$$$\frac{\frac{{R}_{{o}} }{{p}}+\frac{{R}_{{o}} }{{q}}+\frac{{R}_{{o}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}={R}_{{o}} \\ $$$$\Rightarrow{R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$
Commented by mr W last updated on 29/Aug/24
Commented by mr W last updated on 28/Aug/24
the largest sphere under the  upper plane (x/a)+(y/b)+(z/c)=1:  say its radius is R_i , then its center  is at (R_i ,R_i ,R_i ).  we have  (((R_i /a)+(R_i /b)+(R_i /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=−R_i   ⇒R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))
$${the}\:{largest}\:{sphere}\:{under}\:{the} \\ $$$${upper}\:{plane}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}}+\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{z}}}{\boldsymbol{{c}}}=\mathrm{1}: \\ $$$${say}\:{its}\:{radius}\:{is}\:{R}_{{i}} ,\:{then}\:{its}\:{center} \\ $$$${is}\:{at}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right). \\ $$$${we}\:{have} \\ $$$$\frac{\frac{{R}_{{i}} }{{a}}+\frac{{R}_{{i}} }{{b}}+\frac{{R}_{{i}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=−{R}_{{i}} \\ $$$$\Rightarrow{R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$
Commented by mr W last updated on 28/Aug/24
Commented by mr W last updated on 29/Aug/24
let′s ignore the lower plane at first.  as shown previously, the largest   sphere under the upper plane has   the radius R_i  and and its center is   at (R_i ,R_i ,R_i ).  R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))  if the distance from (R_i ,R_i ,R_i ) to  the lower plane is equal to or larger  than R_i , i.e. if  (((R_i /p)+(R_i /q)+(R_i /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))≥R_i , i.e. if  (1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))≤R_i   i.e. if R_o ≤R_i ,  that means if   (1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))  ≥(1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))  then the lower plane is beneath  the sphere with radius R_i  and this   sphere is also the largest sphere  between both planes. that means  in this case  R=R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))    if R_o >R_i , i.e. if  (1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))  <(1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))  then the largest sphere between  both planes is the maximum sphere  from following three cases:    case 1:   center at (u, R_1 ,R_1 ), radius R_1   (((u/a)+(R_1 /b)+(R_1 /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=−R_1   ⇒u=a−aR_1 ((1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))  ⇒u=a−aR_1 ((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))+R_1   ⇒u=a−((aR_1 )/R_i )+R_1   (((u/p)+(R_1 /q)+(R_1 /r)−1)/( (√((1/p^2 )+(1/q^2 )+(1/r^2 )))))=R_1   ⇒u=p−pR_1 ((1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 ))))  ⇒u=p−pR_1 ((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 ))))+R_1   ⇒u=p−((pR_1 )/R_o )+R_1   p−((pR_1 )/R_o )+R_1 =a−((aR_1 )/R_i )+R_1   (R_1 /R_i )(1−(p/a)×(R_i /R_o ))=1−(p/a)  ⇒R_1 =((1−(p/a))/(1−(p/a)×(R_i /R_o )))×R_i   u=a(1−(R_1 /R_i ))+R_1   ⇒u=a(1−((1−(p/a))/(1−(p/a)×(R_i /R_o ))))+R_1  > R_1     similarly  case 2:   center at (R_2 ,v,R_2 ), radius R_2   ⇒R_2 =((1−(q/b))/(1−(q/b)×(R_i /R_o )))×R_i   ⇒v=b(1−((1−(q/b))/(1−(q/b)×(R_i /R_o ))))+R_2  > R_2     case 3:   center at (R_3 ,R_3 ,w), radius R_3   ⇒R_3 =((1−(r/c))/(1−(r/c)×(R_i /R_o )))×R_i   ⇒w=c(1−((1−(r/c))/(1−(r/c)×(R_i /R_o ))))+R_3  > R_3     since (R_i /R_o )<1 and 0<(p/a), (q/b), (r/c)≤1  and f(x)=((1−x)/(1−kx)) is a strctly   decreasing function for 0<k<1 and  0≤x≤1, we get  R=max(R_1 ,R_2 ,R_3 )     =((1−min((p/a),(q/b),(r/c)))/(1−min((p/a),(q/b),(r/c))×(R_i /R_o )))×R_i
$${let}'{s}\:{ignore}\:{the}\:{lower}\:{plane}\:{at}\:{first}. \\ $$$${as}\:{shown}\:{previously},\:{the}\:{largest}\: \\ $$$${sphere}\:{under}\:{the}\:{upper}\:{plane}\:{has}\: \\ $$$${the}\:{radius}\:{R}_{{i}} \:{and}\:{and}\:{its}\:{center}\:{is}\: \\ $$$${at}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right). \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${if}\:{the}\:{distance}\:{from}\:\left({R}_{{i}} ,{R}_{{i}} ,{R}_{{i}} \right)\:{to} \\ $$$${the}\:{lower}\:{plane}\:{is}\:{equal}\:{to}\:{or}\:{larger} \\ $$$${than}\:{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\frac{{R}_{{i}} }{{p}}+\frac{{R}_{{i}} }{{q}}+\frac{{R}_{{i}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}\geqslant{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}\leqslant{R}_{{i}} \\ $$$${i}.{e}.\:{if}\:{R}_{{o}} \leqslant{R}_{{i}} , \\ $$$${that}\:{means}\:{if}\: \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }} \\ $$$$\geqslant\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${then}\:{the}\:{lower}\:{plane}\:{is}\:{beneath} \\ $$$${the}\:{sphere}\:{with}\:{radius}\:{R}_{{i}} \:{and}\:{this}\: \\ $$$${sphere}\:{is}\:{also}\:{the}\:{largest}\:{sphere} \\ $$$${between}\:{both}\:{planes}.\:{that}\:{means} \\ $$$${in}\:{this}\:{case} \\ $$$${R}={R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$$ \\ $$$${if}\:{R}_{{o}} >{R}_{{i}} ,\:{i}.{e}.\:{if} \\ $$$$\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }} \\ $$$$<\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${then}\:{the}\:{largest}\:{sphere}\:{between} \\ $$$${both}\:{planes}\:{is}\:{the}\:{maximum}\:{sphere} \\ $$$${from}\:{following}\:{three}\:{cases}: \\ $$$$ \\ $$$${case}\:\mathrm{1}:\: \\ $$$${center}\:{at}\:\left({u},\:{R}_{\mathrm{1}} ,{R}_{\mathrm{1}} \right),\:{radius}\:{R}_{\mathrm{1}} \\ $$$$\frac{\frac{{u}}{{a}}+\frac{{R}_{\mathrm{1}} }{{b}}+\frac{{R}_{\mathrm{1}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=−{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}−{aR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{u}={a}−{aR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}−\frac{{aR}_{\mathrm{1}} }{{R}_{{i}} }+{R}_{\mathrm{1}} \\ $$$$\frac{\frac{{u}}{{p}}+\frac{{R}_{\mathrm{1}} }{{q}}+\frac{{R}_{\mathrm{1}} }{{r}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}}={R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={p}−{pR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{u}={p}−{pR}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={p}−\frac{{pR}_{\mathrm{1}} }{{R}_{{o}} }+{R}_{\mathrm{1}} \\ $$$${p}−\frac{{pR}_{\mathrm{1}} }{{R}_{{o}} }+{R}_{\mathrm{1}} ={a}−\frac{{aR}_{\mathrm{1}} }{{R}_{{i}} }+{R}_{\mathrm{1}} \\ $$$$\frac{{R}_{\mathrm{1}} }{{R}_{{i}} }\left(\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }\right)=\mathrm{1}−\frac{{p}}{{a}} \\ $$$$\Rightarrow{R}_{\mathrm{1}} =\frac{\mathrm{1}−\frac{{p}}{{a}}}{\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$${u}={a}\left(\mathrm{1}−\frac{{R}_{\mathrm{1}} }{{R}_{{i}} }\right)+{R}_{\mathrm{1}} \\ $$$$\Rightarrow{u}={a}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{p}}{{a}}}{\mathrm{1}−\frac{{p}}{{a}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{1}} \:>\:{R}_{\mathrm{1}} \\ $$$$ \\ $$$${similarly} \\ $$$${case}\:\mathrm{2}:\: \\ $$$${center}\:{at}\:\left({R}_{\mathrm{2}} ,{v},{R}_{\mathrm{2}} \right),\:{radius}\:{R}_{\mathrm{2}} \\ $$$$\Rightarrow{R}_{\mathrm{2}} =\frac{\mathrm{1}−\frac{{q}}{{b}}}{\mathrm{1}−\frac{{q}}{{b}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$$\Rightarrow{v}={b}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{q}}{{b}}}{\mathrm{1}−\frac{{q}}{{b}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{2}} \:>\:{R}_{\mathrm{2}} \\ $$$$ \\ $$$${case}\:\mathrm{3}:\: \\ $$$${center}\:{at}\:\left({R}_{\mathrm{3}} ,{R}_{\mathrm{3}} ,{w}\right),\:{radius}\:{R}_{\mathrm{3}} \\ $$$$\Rightarrow{R}_{\mathrm{3}} =\frac{\mathrm{1}−\frac{{r}}{{c}}}{\mathrm{1}−\frac{{r}}{{c}}×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$$$\Rightarrow{w}={c}\left(\mathrm{1}−\frac{\mathrm{1}−\frac{{r}}{{c}}}{\mathrm{1}−\frac{{r}}{{c}}×\frac{{R}_{{i}} }{{R}_{{o}} }}\right)+{R}_{\mathrm{3}} \:>\:{R}_{\mathrm{3}} \\ $$$$ \\ $$$${since}\:\frac{{R}_{{i}} }{{R}_{{o}} }<\mathrm{1}\:{and}\:\mathrm{0}<\frac{{p}}{{a}},\:\frac{{q}}{{b}},\:\frac{{r}}{{c}}\leqslant\mathrm{1} \\ $$$${and}\:{f}\left({x}\right)=\frac{\mathrm{1}−{x}}{\mathrm{1}−{kx}}\:{is}\:{a}\:{strctly}\: \\ $$$${decreasing}\:{function}\:{for}\:\mathrm{0}<{k}<\mathrm{1}\:{and} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1},\:{we}\:{get} \\ $$$${R}={max}\left({R}_{\mathrm{1}} ,{R}_{\mathrm{2}} ,{R}_{\mathrm{3}} \right) \\ $$$$\:\:\:=\frac{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)}{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)×\frac{{R}_{{i}} }{{R}_{{o}} }}×{R}_{{i}} \\ $$
Commented by mr W last updated on 27/Aug/24
summary:  with   R_i =(1/((1/a)+(1/b)+(1/c)+(√((1/a^2 )+(1/b^2 )+(1/c^2 )))))  R_o =(1/((1/p)+(1/q)+(1/r)−(√((1/p^2 )+(1/q^2 )+(1/r^2 )))))  if R_o ≤R_i , then  R=R_i   if R_o >R_i , then  R=((1−min((p/a),(q/b),(r/c)))/(1−min((p/a),(q/b),(r/c))∙(R_i /R_o )))∙R_i
$${summary}: \\ $$$${with}\: \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}−\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$${if}\:{R}_{{o}} \leqslant{R}_{{i}} ,\:{then} \\ $$$${R}={R}_{{i}} \\ $$$${if}\:{R}_{{o}} >{R}_{{i}} ,\:{then} \\ $$$${R}=\frac{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)}{\mathrm{1}−{min}\left(\frac{{p}}{{a}},\frac{{q}}{{b}},\frac{{r}}{{c}}\right)\centerdot\frac{{R}_{{i}} }{{R}_{{o}} }}\centerdot{R}_{{i}} \\ $$
Commented by mr W last updated on 27/Aug/24
formula used:  distance from point to plane in space
$${formula}\:{used}: \\ $$$${distance}\:{from}\:{point}\:{to}\:{plane}\:{in}\:{space} \\ $$
Commented by mr W last updated on 27/Aug/24
Commented by ajfour last updated on 27/Aug/24
immense ! i will look carefully. get  the idea very well. great thing for  question treasure.
$${immense}\:!\:{i}\:{will}\:{look}\:{carefully}.\:{get} \\ $$$${the}\:{idea}\:{very}\:{well}.\:{great}\:{thing}\:{for} \\ $$$${question}\:{treasure}. \\ $$
Commented by mr W last updated on 27/Aug/24
thanks for checking sir!
$${thanks}\:{for}\:{checking}\:{sir}! \\ $$
Commented by mr W last updated on 27/Aug/24
example 1:  upper plane (x/6)+(y/9)+(z/(10))=1  lower plane (x/5)+(y/3)+(z/4)=1  a=6, b=9, c=10  p=5, q=3, r=4  (p/a)=(5/6), (q/b)=(3/9)=(1/3), (r/c)=(4/(10))=(2/5)   ⇒min=(q/b)=(1/3)  R_i =(1/((1/6)+(1/9)+(1/(10))+(√((1/6^2 )+(1/9^2 )+(1/(10^2 ))))))        =((102−3(√(406)))/(25))≈1.662  R_o =(1/((1/5)+(1/3)+(1/4)−(√((1/5^2 )+(1/3^2 )+(1/4^2 )))))       =((47+(√(769)))/(24))≈3.114  R_o >R_i    R=((1−(1/3))/(1−(1/3)×((24×(102−3(√(406))))/(25×(47+(√(769)))))))×((102−3(√(406)))/(25))      ≈1.348
$${example}\:\mathrm{1}: \\ $$$${upper}\:{plane}\:\frac{{x}}{\mathrm{6}}+\frac{{y}}{\mathrm{9}}+\frac{{z}}{\mathrm{10}}=\mathrm{1} \\ $$$${lower}\:{plane}\:\frac{{x}}{\mathrm{5}}+\frac{{y}}{\mathrm{3}}+\frac{{z}}{\mathrm{4}}=\mathrm{1} \\ $$$${a}=\mathrm{6},\:{b}=\mathrm{9},\:{c}=\mathrm{10} \\ $$$${p}=\mathrm{5},\:{q}=\mathrm{3},\:{r}=\mathrm{4} \\ $$$$\frac{{p}}{{a}}=\frac{\mathrm{5}}{\mathrm{6}},\:\frac{{q}}{{b}}=\frac{\mathrm{3}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}},\:\frac{{r}}{{c}}=\frac{\mathrm{4}}{\mathrm{10}}=\frac{\mathrm{2}}{\mathrm{5}}\: \\ $$$$\Rightarrow{min}=\frac{{q}}{{b}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${R}_{{i}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+\sqrt{\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}}{\mathrm{25}}\approx\mathrm{1}.\mathrm{662} \\ $$$${R}_{{o}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{47}+\sqrt{\mathrm{769}}}{\mathrm{24}}\approx\mathrm{3}.\mathrm{114} \\ $$$${R}_{{o}} >{R}_{{i}} \: \\ $$$${R}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{24}×\left(\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}\right)}{\mathrm{25}×\left(\mathrm{47}+\sqrt{\mathrm{769}}\right)}}×\frac{\mathrm{102}−\mathrm{3}\sqrt{\mathrm{406}}}{\mathrm{25}} \\ $$$$\:\:\:\:\approx\mathrm{1}.\mathrm{348} \\ $$
Commented by behi834171 last updated on 27/Aug/24
mathmatic is very beautiful.  your Q′s & A′s ,are more than   beautiful!  They are amazing sir mrW.  Thanks god for being of you in this forum.
$${mathmatic}\:{is}\:{very}\:{beautiful}. \\ $$$${your}\:{Q}'{s}\:\&\:{A}'{s}\:,{are}\:{more}\:{than}\: \\ $$$${beautiful}! \\ $$$${They}\:{are}\:{amazing}\:{sir}\:{mrW}. \\ $$$${Thanks}\:{god}\:{for}\:{being}\:{of}\:{you}\:{in}\:{this}\:{forum}. \\ $$
Commented by mr W last updated on 27/Aug/24
thanks back to you, father from Behi,   for your appreciation and support!
$${thanks}\:{back}\:{to}\:{you},\:{father}\:{from}\:{Behi},\: \\ $$$${for}\:{your}\:{appreciation}\:{and}\:{support}! \\ $$
Commented by necx122 last updated on 27/Aug/24
Mr. Behi its been a really long time. Feels  good to see your post here again.
$${Mr}.\:{Behi}\:{its}\:{been}\:{a}\:{really}\:{long}\:{time}.\:{Feels} \\ $$$${good}\:{to}\:{see}\:{your}\:{post}\:{here}\:{again}. \\ $$
Commented by ajfour last updated on 27/Aug/24
yes so glad for your remark bsck on this forum.
$${yes}\:{so}\:{glad}\:{for}\:{your}\:{remark}\:{bsck}\:{on}\:{this}\:{forum}. \\ $$
Commented by behi834171 last updated on 28/Aug/24
Red heart to you dear mr.W
$${Red}\:{heart}\:{to}\:{you}\:{dear}\:{mr}.{W} \\ $$
Commented by behi834171 last updated on 28/Aug/24
Thank you so much sir with red herat!
$${Thank}\:{you}\:{so}\:{much}\:{sir}\:{with}\:{red}\:{herat}! \\ $$
Commented by behi834171 last updated on 28/Aug/24
so happy for see you  again sir ajfour   and red heart to you!
$${so}\:{happy}\:{for}\:{see}\:{you}\:\:{again}\:{sir}\:{ajfour} \\ $$$$\:{and}\:{red}\:{heart}\:{to}\:{you}! \\ $$
Commented by BHOOPENDRA last updated on 01/Sep/24
great Mr.W
$${great}\:{Mr}.{W} \\ $$

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