行列式在空间解析几何中的应用 空间解析几何中最基本的方法是坐标法,即建立一个坐标系,使得点用有序 实数组(即坐标)表示,从而图形可以用方程来表示,一些相关的几何量可以用向 量以及向量的内积,外积,混合积来表示,而这些与行列式密切相关,因此,几何 中的一些问题可以通过行列式找到更加快捷,简便的解决方法 求面积,体积的问题 引理1已知空间两向量,a=(a1,a2,a3),B=(b1,b2,b3),则a与B的外积 a2 a3 31 b2b3'b3b1'|b1b2 引理2已知空间三向量a=(a1,a2,a3),3=(b1,b2,b3),=(c1,c2,3),则混 合积 a1 a2 a (a,B,)=b1b2b3 定理3已知空间中三角形的三个顶点为A(x2,,z)(i=1,2,3),则△A1A2A3 的面积为 1 △A1A2432y3-y123-21 2-z1x2-1+/2-x1y- r3-x133-y1 证明:由于向量A1A2与A143的外积A1A2×A143=A1421A3in,(为 向量A1A2与A1A3的夹角),因此S△A1AA4=A142×A1A3,利用引理1,可得 结论 推论4在平面直角坐标系下,设三点为A(x1,v)(=1,2,3),则 △A1A2A 11x2-1y-0的绝对值 2|x3-x13-y y1 即S△A1A2A3=x2y1的绝对值
✁✂✄☎✆✝✞✟✠✡☛☞✌ ✍✎✏✑✒✓✔✕✖✗✘✙✚✛✜✢✚✣✤✥✦✧★✜✢✩✣✪✫✬✭✮✯ ✰✱✲ (✤✜✢) ✳✴✣✵✶✷✸✹✺✭✙✻✼✳✴✣✧✽✾✿✘✒✓❀✹✺✭❁ ❀✺❂❁❀✘❃❄✣❅❄✣❆❇❄✼✳✴✣✶❈✽❉❊❋●❍■✾✿✣❏❑✣✒✓ ✔✘✧✽▲▼✹✺◆❖❊❋●P◗❘❙❚❯✣❱❲✘✏❳✙✚❨ ❩❨❬❭❪✣❫❪❴❵❛ ❜❝ 1 ❞ ❡✍✎❢❁❀✣ α = (a1, a2, a3),β = (b1, b2, b3), ❣ α ❉ β ✘❅❄ α × β = a2 a3 b2 b3 , a3 a1 b3 b1 , a1 a2 b1 b2 ! . ❜❝ 2 ❞ ❡✍✎❤❁❀ α = (a1, a2, a3),β = (b1, b2, b3) ,γ = (c1, c2, c3), ❣ ❆ ❇❄ (α, β, γ) = a1 a2 a3 b1 b2 b3 c1 c2 c3 . ✐❝ 3 ❞ ❡✍✎✔❤❥✸✘❤★❦✬❧ Ai(xi , yi , zi)(i = 1, 2, 3), ❣ 4A1A2A3 ✘♠❄❧ S4A1A2A3 = 1 2 vuut y1 − y2 z2 − z1 y3 − y1 z3 − z1 2 + z2 − z1 x2 − x1 z3 − z1 x3 − x1 2 + x2 − x1 y2 − y1 x3 − x1 y3 − y1 2 . ♥♦♣qr❁❀ −−−→ A1A2 ❉ −−−→ A1A3 ✘❅❄ −−−→ A1A2× −−−→ A1A3= | −−−→ A1A2||−−−→ A1A3|sin θ, (θ ❧ ❁❀ −−−→ A1A2 ❉ −−−→ A1A3 ✘s❥ ), ❏❑ S4A1A2A3= 1 2 k −−−→ A1A2× −−−→ A1A3k, t ✭✉✈ 1, ✹✫ ✇①❨ ②③ 4 ④⑤♠⑥❥✜✢✩⑦✣⑧❤✬❧ Ai(xi , yi)(i = 1, 2, 3) , ❣ S4A1A2A3 = 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 ✘⑨⑩❶, ✤ S4A1A2A3 = 1 2 x1 y1 1 x2 y2 1 x3 y3 1 ✘⑨⑩❶❨ 1
证明:由定理3易得S△A=|2-nD=m的绝对值又因为 0 1y-3/1 y11 所以,S△A4= 的绝对值 下面从推论4三角形面积的行列式公式入手,通过几条不同的途径,进一步 阐述三角形面积的几种行列式公式 (1)已知三个顶点的情形 将坐标平面看作复平面,△A1A243的顶点用复数k=xk+ik表示,(k= 1,2,3)由于 I+iy1 y1 1 ly1 C2 92 T2+ig2 92 1=1 z2 -ig2 1 3+iy3331 3 r1x1+ T2 92 1=i a2 r2+ig2 3 y3 33+ 2y3 由(1),(2)可得 其中买是xk的共轭复数,所以 △A1A2A3 2z1的模 4 2)已知三边的情形 设在直角坐标系下,S△A1A2A3的三边方程为akx+bky+ck,(k=1,2,3),应 用 cramer法则,将交点坐标用系数表示出来,代入推论4中公式,易得 2
♥♦♣q❷✈ 3, ❸ ✫ S4A1A2A3 = 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 ✘⑨⑩❶❹❏❧ 1 2 x2 − x1 y2 − y1 x3 − x1 y3 − y1 = 1 2 0 0 1 x2 − x1 y2 − y1 1 x3 − x1 y3 − y1 1 = 1 2 x1 y1 1 x2 y2 1 x3 y3 1 , ❺✺✣ S4A1A2A3 = 1 2 x1 y1 1 x2 y2 1 x3 y3 1 ✘⑨⑩❶❨ ⑦♠✵❻① 4 ❤❥✸♠❄✘❊❋●❼●❽❾✣◆❖✒❿➀ ➁✘➂➃✣➄✧➅ ➆➇❤❥✸♠❄✘✒➈❊❋●❼●❨ (1) ➉➊➋➌➍➎❴➏➐ ➑✜✢⑤ ♠➒➓➔⑤ ♠✣ 4A1A2A3 ✘❦✬✭➔✱ zk = xk + iyk ✳✴✣ (k = 1, 2, 3) qr x1 y1 1 x2 y2 1 x3 y3 1 = x1 + iy1 y1 1 x2 + iy2 y2 1 x3 + iy3 y3 1 = i z1 −iy1 1 z2 −iy2 1 z3 −iy3 1 (1) x1 y1 1 x2 y2 1 x3 y3 1 = i x1 x1 + iy1 1 x2 x2 + iy2 1 x3 x3 + iy3 1 = i z1 x1 1 z2 x2 1 z3 x3 1 (2) q (1),(2) ✹✫ x1 y1 1 x2 y2 1 x3 y3 1 = i 2 z1 z1 1 z2 z2 1 z3 z3 1 , →✔ zk ✛ zk ✘➣↔➔✱✣❺✺ S4A1A2A3 = i 4 z1 z1 1 z2 z2 1 z3 z3 1 ✘↕. (2) ➉➊➋➙❴➏➐ ⑧ ④ ⑥❥✜✢✩⑦✣ S4A1A2A3 ✘❤➛✙✻❧ akx + bky + ck ,(k = 1, 2, 3), ➜ ✭ cramer ✚ ❣ ✣➑➝✬✜✢✭✩✱ ✳✴➞✼✣➟❽❻① 4 ✔❼●✣❸ ✫ 2
b1 b2 b1 b2 a1 b C1 的绝对值.记|4为a2b2c2, 1 b1\\为 a1 b a2 b2 A1412413 A 1 A22 A A31A432A 其中A是|A中第k行,j列处元素的代数余子式,则 a1A1+b1A12+c1413 a2A21+b2A22+c2A A31+b3A3g+(34 A11A12A1 于是若4≠0,则A21A2A3=4P.因此, A31A32A3 b1 S△AA2=Ta1h22a2b22的绝对值 b2 b 3 b3 (3)已知三边长的情形 仍设三顶点为Ak(xk,),ak;表示点Ak与A的距离,并令(ak,a)=xkx+ 孙,(k,j=1,2,3),则a3,=(ak,ak)+(ay,a)-2(ak,a) 于是
S4A1A2A3= 1 2 a1 b1 a2 b2 a1 b1 a3 b3 a2 b2 a3 b3 b2 c2 b3 c3 − a2 c2 a3 c3 a2 b2 a3 c3 − b1 c1 b3 c3 a1 c1 a3 c3 − a1 b1 a3 b3 b1 c1 b2 c2 − a1 c1 a2 c2 a1 b1 a2 b2 ✘⑨⑩❶❨➠ |A| ❧ a1 b1 c1 a2 b2 c2 a3 b3 c3 , b2 c2 b3 c3 − a2 c2 a3 c3 a2 b2 a3 c3 − b1 c1 b3 c3 a1 c1 a3 c3 − a1 b1 a3 b3 b1 c1 b2 c2 − a1 c1 a2 c2 a1 b1 a2 b2 ❧ A11 A12 A13 A21 A22 A23 A31 A32 A33 , →✔ Akj ✛ |A| ✔➡ k ❊✣ j ❋➢➤➥✘➟✱➦➧●✣❣ a1A11 + b1A12 + c1A13 0 0 0 a2A21 + b2A22 + c2A23 0 0 0 a3A31 + b3A32 + c3A33 = |A| 3 , r✛➨ |A| 6= 0, ❣ A11 A12 A13 A21 A22 A23 A31 A32 A33 =|A| 2 . ❏❑✣ S4A1A2A3 = 1 2 a1 b1 a2 b2 a1 b1 a3 b3 a2 b2 a3 b3 a1 b1 c1 a2 b2 c2 a3 b3 c3 2 ✘⑨⑩❶. (3) ➉➊➋➙➩❴➏➐ ➫⑧❤❦✬❧ Ak(xk, yk),akj ✳✴✬ Ak ❉ Aj ✘➭➯✣➲➳ (αk, αj )=xkxj + ykyj ,(k, j = 1, 2, 3), ❣ α 2 kj= (αk, αk) + (αj , αj ) − 2(αk, αj ) r✛✣ 3
S△A1A2A3 1yy33 v31111 1 (a1,a2)+1( (a2,a1)+1(a2,a2)+1(a2,a3)+1 4|(34)+1(a32)+1(a303)+1 41(a2,a1)(a2a2)(a 1(a3,a1)(a3,a2)(as3,a3) 0111 161 0a2 0 因此,用边长表示的三角形面积公式为 D(A1,A2,A3) 这里 0111 D(A1,A2,A3) 220 叫做点集{A1,A2,Aa3}的 Cayley- Menger行列式 由推论4(2)式可得 推论5设在平面直角坐标系下,四边形A1A2A3A4四个顶点的坐标为(xk,k) (k=1,2,3,4),则它的面积 y1 S0A142434= r2y01 3为811的绝对值.用复数表示,命x1=+i0mk= 01 1,2,3,4),则 21z11 SOAlA2A3A4 =4 3211的模 4
S4A1A2A3 2 = 1 4 x1 y1 1 x2 y2 1 x3 y3 1 x1 x2 x3 y1 y2 y3 1 1 1 = 1 4 (α1, α1) + 1 (α1, α2) + 1 (α1, α3) + 1 (α2, α1) + 1 (α2, α2) + 1 (α2, α3) + 1 (α3, α1) + 1 (α3, α2) + 1 (α3, α3) + 1 = − 1 4 0 1 1 1 1 (α1, α1) (α1, α2) (α1, α3) 1 (α2, α1) (α2, α2) (α2, α3) 1 (α3, α1) (α3, α2) (α3, α3) = − 1 16 0 1 1 1 1 0 a 2 12 a 2 13 1 a 2 21 0 a 2 23 1 a 2 31 a 2 32 0 . ❏❑✣✭➛➵✳✴✘❤❥✸♠❄❼●❧ S4A1A2A3 = − 1 16 D(A1, A2, A3), ❈➸ D(A1, A2, A3) = 0 1 1 1 1 0 a 2 12 a 2 13 1 a 2 21 0 a 2 23 1 a 2 31 a 2 32 0 ➺➻✬➼ {A1, A2, A3} ✘ Cayley-Menger ❊❋●❨ q❻① 4(2) ●✹✫ ②③ 5 ⑧ ④⑤♠⑥❥✜✢✩⑦✣➽➛✸ A1A2A3A4 ➽★❦✬✘✜✢❧ (xk, yk) ,(k = 1, 2, 3, 4), ❣➾✘♠❄ S♦A1A2A3A4= 1 2 x1 y1 1 1 x2 y2 0 1 x3 y3 1 1 x4 y4 0 1 ✘⑨⑩❶❨✭➔✱ ✳✴✣➚ zk = xk + iyk,(k = 1, 2, 3, 4), ❣ S♦A1A2A3A4 = i 4 z1 z1 1 1 z2 z2 0 1 z3 z3 1 1 z4 z4 0 1 ✘↕❨ 4
定理6以不在同一平面上的四点A(x2,y,z)(i=1,2,3,4)为顶点的平行六 y121 面体的体积为V= 的绝对值 3势 证明:易见 A1A2=(x2-x1,y-y1,22-x1) A1A3=(x3-x1,y8-91,23-21) A144 以这三个向量为棱的平行六面体的体积为这三个向量的混合积(A1A2A1A3,A14) 的绝对值 2-192-y122-21 (A1A241A3,A1A4)=x3-x1-23-21 1y2-y12-21 y3-3/123-21 4-1y4-y/124-21 结论得证 推论7以A(x2,y,z),(i=1,2,3,4)为顶点的四面体的体积为 y1 V 2y22 的绝对值 共点共线与共面的条件 定理8在平面直角坐标系中,三点A1(x1,v),(=1,2,3)共线的充要条件是 xxx y1 y2 证明:由推论4,易得 定理9空间四点A(xm,)(=12,3,4)共面的充要条件是2m21=0
✐❝ 6 ✺➀ ④ ➁✧ ⑤ ♠➪✘ ➽✬ Ai(xi , yi , zi)(i = 1, 2, 3, 4) ❧❦✬✘⑤ ❊➶ ♠➹✘➹❄❧ V = x1 y1 z1 1 x2 y2 z2 1 x3 y3 z3 1 x4 y4 z4 1 ✘⑨⑩❶❨ ♥♦♣❸➘ −−−→ A1A2 = (x2 − x1, y2 − y1, z2 − z1) −−−→ A1A3 = (x3 − x1, y3 − y1, z3 − z1) −−−→ A1A4 = (x4 − x1, y4 − y1, z4 − z1) ✺❈❤★❁❀❧➴✘⑤ ❊➶♠➹✘➹❄❧❈❤★❁❀✘❆❇❄ ( −−−→ A1A2, −−−→ A1A3, −−−→ A1A4) ✘⑨⑩❶❨ ( −−−→ A1A2, −−−→ A1A3, −−−→ A1A4) = x2 − x1 y2 − y1 z2 − z1 x3 − x1 y3 − y1 z3 − z1 x4 − x1 y4 − y1 z4 − z1 = 0 0 0 1 x2 − x1 y2 − y1 z2 − z1 1 x3 − x1 y3 − y1 z3 − z1 1 x4 − x1 y4 − y1 z4 − z1 1 = − x1 y1 z1 1 x2 y2 z2 1 x3 y3 z3 1 x4 y4 z4 1 , ✇①✫♥❨ ②③ 7 ✺ Ai(xi , yi , zi),(i=1,2,3,4) ❧❦✬✘➽♠➹✘➹❄❧ V = 1 6 x1 y1 z1 1 x2 y2 z2 1 x3 y3 z3 1 x4 y4 z4 1 ✘⑨⑩❶❨ ➷❨➬➎ ➬➮➱➬❭❴✃❐ ✐❝ 8 ④⑤♠⑥❥✜✢✩✔✣❤✬ Ai(xi , yi),(i = 1, 2, 3) ➣❒✘❮❰❿Ï✛ x1 y1 1 x2 y2 1 x3 y3 1 = 0. ♥♦♣q❻① 4, ❸ ✫❨ ✐❝ 9 ✍✎➽✬ Ai(xi , yi , zi)(i = 1, 2, 3, 4) ➣♠✘❮❰❿Ï✛ x1 y1 z1 1 x2 y2 z2 1 x3 y3 z3 1 x4 y4 z4 1 =0 5
证明:由定理6易得. 定理10三个平面A1+By+C1z+D2=0(i=1,2,3)交于一点的充要条件 是 A1 BI cl A2B2C2≠0 A B3 C3 证明:由 Cramer法则易得. 三.空间直线的方程 定理11空间上不在同一直线上的三个点A(x2vn,x)(=1,2,3)所确定的平 面方程为 y1z11 0 y222 证明:易看出:上式为关于x,y,z的线形方程,设为f(x,y,x)=Ax+By+ Cz+D=0,因为A1,A2,A3不在同一直线上,所以A,B,C不全为零,因此它是 个平面方程.再者A1,A2,A3在此平面上,因为 2,y/2,2 定理得证 推论12通过点(a,b,c)与直线=m0=5a的平面方程为 y l+ 证明:易见( )与(+ v0,n+20)是在直线 上不同的两个点,由定理11,即得本推论的结果 6
♥♦♣q❷✈ 6 ❸ ✫❨ ✐❝ 10 ❤★ ⑤ ♠ Aix + Biy + Ciz + Di = 0(i = 1, 2, 3) ➝r✧✬✘❮❰❿Ï ✛ A1 B1 C1 A2 B2 C2 A3 B3 C3 6= 0. ♥♦♣q Cramer ✚ ❣❸✫❨ ➋ ❨ÐÑÒ➮❴ÓÔ ✐❝ 11 ✍✎➪➀ ④ ➁✧⑥❒➪✘❤★✬ Ai(xi , yi , zi)(i = 1, 2, 3) ❺Õ❷✘ ⑤ ♠✙✻❧ x y z 1 x1 y1 z1 1 x2 y2 z2 1 x3 y3 z3 1 = 0 ♥♦♣❸ ➒ ➞♣➪●❧✿r x, y, z ✘❒✸✙✻✣⑧❧ f(x, y, z) = Ax + By + Cz + D = 0, ❏❧ A1, A2, A3 ➀ ④ ➁✧⑥❒➪✣❺✺ A, B, C ➀Ö❧×✣❏❑ ➾ ✛ ✧★⑤ ♠✙✻❨ØÙ A1, A2, A3 ④ ❑ ⑤ ♠➪✣❏❧ f(x1, y1, z1) = f(x2, y2, z2) = f(x3, y3, z3) = 0 . ❷✈✫♥❨ ②③ 12 ◆❖✬ (a, b, c) ❉⑥❒ x−x0 l = y−y0 m = z−z0 n ✘ ⑤ ♠✙✻❧ x y z 1 a b c 1 x0 y0 z0 1 l + x0 m + y0 n + z0 1 = 0 . ♥♦♣❸➘ (x0, y0, z0) ❉ (l + x0, m + y0, n + z0) ✛ ④ ⑥❒ x−x0 l = y−y0 m = z−z0 n ➪➀➁✘❢★✬✣q❷✈ 11, ✤✫✗❻①✘✇Ú❨ 6
定理13过不在同一直线上三点(x1,y),(x2,y2),(x3,93)的圆的方程为 2222 yy ++++ y1-1y1 y2-29 t3 y 证明:依第一行展开上面行列式,得Ax2+Ay2+Bx+Cy+D=0,这是 个圆的方程,又从行列式容易看出(x1,y),(x2,y2),(x3,y3)均满足此圆的方程 (王忠梅编写) 7
✐❝ 13 ❖➀ ④ ➁✧⑥❒➪❤✬ (x1, y1),(x2, y2),(x3, y3) ✘Û✘✙✻❧ x 2 + y 2 x y 1 x1 2 + y1 2 x1 y1 1 x2 2 + y2 2 x2 y2 1 x3 2 + y3 2 x3 y3 1 = 0 . ♥♦♣Ü➡✧❊ÝÞ➪♠❊❋●✣✫ Ax2 + Ay2 + Bx + Cy + D = 0, ❈✛✧ ★Û✘✙✻✣❹✵❊❋●ß❸ ➒ ➞ (x1, y1),(x2, y2),(x3, y3) àáâ❑Û✘✙✻❨ (ãäå æç) 7