Question Number 149428 by mr W last updated on 06/Aug/21

Commented bymr W last updated on 06/Aug/21

three non−collinear points in space are given with coordinates A(x_1 ,y_1 ,z_1 ) B(x_2 ,y_2 ,z_2 ) C(x_3 ,y_3 ,z_3 ) find the coordinates of the center S and the radius R of the circumcircle.

Commented byajfour last updated on 06/Aug/21

sir, (apart from this); sir try plotting z(x,y)={(x^2 −1)^3 +(y^2 −1)^3 }^2 see all views, change scale, wonderful (i find)...

Commented bymr W last updated on 06/Aug/21

Commented byajfour last updated on 06/Aug/21

Answered by ajfour last updated on 06/Aug/21

A(a), B(b), C(c) plane ABC (s−a).((b−c)×(a−c))=0 let S(s). n^� =(((b−c)×(a−c))/(∣(b−c)×(a−c)∣)) let s=(((b+c))/2)+((∣b−c∣)/(2tan A)){((n^� ×(b−c))/( ∣b−c∣))} ∣s−a∣=∣s−b∣=∣s−c∣=R

Commented bymr W last updated on 06/Aug/21

thanks sir! i′ll also try to use vectors.

Commented byajfour last updated on 06/Aug/21

let A(5,2,3) ; B(3,5,5) ; C(2,3,8) n^� =(((i+2j−3k)×(2i−3j−2k))/(∣(i+2j−3k)×(2i−3j−2k)∣)) =((−3k+2j−4k+4i+6j−9i)/( (√(25+64+49)))) =((−5i+8j−7k)/( (√(138)))) s=(((5i+8j+13k))/2) +((∣i+2j−3k∣)/(2tan A))(((−5i+8j−7k)/( (√(138))))×((i+2j−3k)/( (√(14))))) let D be mid-point of AB=((a+b)/2) CD=((a+b−2c)/2)=((4i+j−8k)/2) ((AB)/2)=((−2i+3j+2k)/2) tan A=((CD)/((AB)/2))=((√(16+1+64))/( (√(4+9+4)))) =(9/( (√(17)))) s=((5i+8j+13k)/2) +((√(17))/(18(√(138))))(−10k−15j−8k−24i −7j+14i) s=((5i+8j+13k)/2) +(((√(17))(−10i−22j−18k))/(18(√(138)))) R=∣s−a∣

Commented bymr W last updated on 07/Aug/21

i got it! it′s a great idea with (1/(tan A))!

Commented byajfour last updated on 07/Aug/21

Answered by mr W last updated on 07/Aug/21

Commented bymr W last updated on 07/Aug/21

let a=OA^(→) , b=OB^(→) , c=OC^(→) , s=OS^(→) let p=AB^(→) , q=BC^(→) , r=CA^(→) normal of plane ABC: n=((r×q)/(∣r×q∣)) D=midpoint of AB OD^(→) =((a+b)/2) ∣DS^(→) ∣=((∣AB^(→) ∣)/2)tan φ=((∣p∣)/(2 tan ∠C)) DS^(→) =∣DS^(→) ∣((p×n)/(∣p∣))=((∣p∣)/(2 tan ∠C))×((p×(r×q))/(∣p∣∣r×q∣)) DS^(→) =(1/(tan ∠C))×((p×(r×q))/(2∣r×q∣)) s=OD^(→) +DS^(→) s=((a+b)/2)+(1/(tan ∠C))×((p×(r×q))/(2∣r×q∣)) cos C=−((q∙r)/(∣q∣∣r∣)) sin C=((∣q×r∣)/(∣q∣∣r∣)) (1/(tan C))=−((q∙r)/(∣q×r∣)) s=((a+b)/2)−((q∙r)/(∣q×r∣))×((p×(r×q))/(2∣r×q∣)) ⇒s=(1/2){a+b+((q∙r)/(∣q×r∣^2 )) p×(q×r)} ⇒R=∣s−a∣=∣s−b∣=∣s−c∣ or R=((∣p∣)/(2 sin ∠C))=((∣p∣∣q∣∣r∣)/(2∣q×r∣)) example: A(5,2,3), B(3,5,5), C(2,3,8) p=(−2,3,2) q=(−1,−2,3) r=(3,−1,−5) q∙r=−3+2−15=−16 q×r= determinant (((−1),(−2),3),(3,(−1),(−5)))=(13,4,7) ∣q×r∣^2 =13^2 +4^2 +7^2 =234 p×(q×r)= determinant (((−2),3,2),((13),4,7))=(13,40,−47) s=(1/2){(5,2,3)+(3,5,5)−((16)/(234))(13,40,−47)} =(((32)/9),((499)/(234)),((656)/(117))) R^2 =∣AS^(→) ∣^2 =(((32)/9)−5)^2 +(((499)/(234))−2)^2 +(((656)/(117))−3)^2 =((4165)/(468)) ✓ R^2 =∣BS^(→) ∣^2 =(((32)/9)−3)^2 +(((499)/(234))−5)^2 +(((656)/(117))−5)^2 =((4165)/(468)) ✓ R^2 =∣CS^(→) ∣^2 =(((32)/9)−2)^2 +(((499)/(234))−3)^2 +(((656)/(117))−8)^2 =((4165)/(468)) ✓ R=(√((4165)/(468)))=((7(√(1105)))/(78))

Commented bymr W last updated on 07/Aug/21

a special case A(u,0,0) B(0,v,0) C(0,0,w) p=(−u,v,0) q=(0,−v,w) r=(u,0,−w) q∙r=−w^2 q×r= determinant ((0,(−v),w),(u,0,(−w)))=(vw,wu,uv) ∣q×r∣^2 =u^2 v^2 +v^2 w^2 +w^2 u^2 p×(q×r)= determinant (((−u),v,0),((vw),(wu),(uv)))=(uv^2 ,u^2 v,−w(u^2 +v^2 )) s=(1/2){(u,0,0)+(0,v,0)−(w^2 /(u^2 v^2 +v^2 w^2 +w^2 u^2 )) (uv^2 ,u^2 v,−w(u^2 +v^2 ))} =(1/2){u−((uv^2 w^2 )/(u^2 v^2 +v^2 w^2 +w^2 u^2 )),v−((u^2 vw^2 )/(u^2 v^2 +v^2 w^2 +w^2 u^2 )),((w^3 (u^2 +v^2 ))/(u^2 v^2 +v^2 w^2 +w^2 u^2 ))} =(1/2){((u^3 (v^2 +w^2 ))/(u^2 v^2 +v^2 w^2 +w^2 u^2 )),((v^3 (w^2 +u^2 ))/(u^2 v^2 +v^2 w^2 +w^2 u^2 )),((w^3 (u^2 +v^2 ))/(u^2 v^2 +v^2 w^2 +w^2 u^2 ))} ∣p∣=(√(u^2 +v^2 )) ∣q∣=(√(v^2 +w^2 )) ∣r∣=(√(w^2 +u^2 )) ∣q×r∣=(√(u^2 v^2 +v^2 w^2 +w^2 u^2 )) ⇒R=(1/2)(√(((u^2 +v^2 )(v^2 +w^2 )(w^2 +u^2 ))/(u^2 v^2 +v^2 w^2 +w^2 u^2 )))