gongxiaoqing@hotmail
DESCRIPTION
浙江省精品课程. 高 等 数 学. 第八章 空间解析几何与向量代数. 龚小庆. [email protected]. B. A. 以 A 为起点, B 为终点的向量, 记为 AB , , a. 向量 AB 的大小叫做 向量的模 . 记为 | AB | 或. 模为 0 的向量称为 零向量 . 记为 0 或 , 它的方向可以看作是任意的. 第一节 向量及其线性运算. 一、向量概念. 1. 向量 : 既有大小 , 又有方向的量 , 称为 向量 ( 或 矢量 ). - PowerPoint PPT PresentationTRANSCRIPT
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1. : , , ()2. : ., .: 1.
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3. : , . .
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1 .(1) (2)
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.(1) : (2) ::
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.(2) .:
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(i) .(ii) .
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21. : . : > 0, < 0, = 0, 2. :(1) :(2) :
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1. ozxy x() y()z(), O.
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2. . , , xoy. yozzox, .
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.RQP< : MM (x, y, z)
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(1) Myoz, x = 0; zox, y = 0; xoy, z = 0.(2) M x , y = z = 0 y , x = z = 0 z , x = y = 0(3) (+, +, +) (, +, +) (, , +) (, , ) (+, , ) (+, , +) (+, +, ) (, +, ):
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M1(x1, y1, z1), M2(x2, y2, z2) d 2 = | M1 M2 |2 = |M1N |2+ |NM2 |2= |P1 P2 |2 + |Q1 Q 2 |2 + |R1 R 2 |2 = |M1P |2 + |PN |2 +|NM2 |2= (x2x1)2 + (y2y1)2 + (z2z1)2
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:: M(x, y, z) O(0,0, 0)
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1: M1(4, 3, 1), M2(7, 1, 2), M3(5, 2, 3).: |M2 M3 | = |M3 M1 |, M1 M2 M3 .
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2 z A(4, 1, 7) B(3, 5, 2). M(0, 0, z) |MA| = |MB|.::
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i, j, kx, y, z, .= xi + yj + zk
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M1 (x1, y1 , z1), M2 (x2, y2 , z2)= (x2 i+ y2 j + z2 k) (x1 i + y1 j + z1 k) = (x2 x1) i + (y2 y1) j + (z2 z1) k (2) a = {x2 x1 , y2 y1 , z2 z1} a ax = x2 x1 , ay = y2 y1 , az = z2 z1 a , a.
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a ={ax , ay , az}, b ={bx , by , bz}, (1) a b = {ax bx , ay by , az bz }(2) a = {ax , ay , az}: (1) a + b = (ax i + ay j+ az k) +(bxi + by j+ bz k)= (ax i + bxi ) +(ay j+ by j) + (az k + bz k)= (ax + bx) i + (ay+ by) j + (az+ bz ) k
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. a ={ax , ay , az}, b ={bx , by , bz}, ax =bx, ay =by, az =bz, a // b(3)
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zxyoABMM:
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1.
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(2): a x, y, z , , , a .(3): cos, cos, cos, .(4) a ={ax, ay, az,}
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(4)(5)
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(5)cos2 +cos2 +cos2 = 1(6)eaa= {cos , cos , cos }(7)
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3. M1(2, 2, )M2(1, 3, 0). M1 M2, .
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(1) Au, Au,uA'Au.
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(2) .u.
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3 . PrjuAB = | AB |cos BBAAuuB
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2.:3: .
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: F, S , .: , , . W=|F |cos |S | = |F | |S | cos
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1. : ab, ,|a||b|cos ab, a b.: a b = |a| |b| cos a 0, | b | cos = Prjab b 0, | a |cos = Prjba a b = |a| Prjab = |b| Prjba: a a = | a |2: i i = j j = k k = 1
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2. (1) a b = b a (2) (a + b) c = a c + b c(3) : ( a) b = a ( b) = (a b)
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(4) a , b a b = 0: : a b, : a b = | a | |b |cos =0; a 0, b 0,: cos =0 , a b: ijk , i j = j k = i k = 0
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1: , | c |2 = | a |2 + | b |2 2 | a | | b |cos: c = a b , | c |2 = | a b |2 = (a b) (a b)= a a + b b 2 a b= | a |2 + | b |2 2 | a | | b |cos| c |2 = | a |2 + | b |2 2 | a | | b |cos:
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3. a ={ax, ay , az}, b = {bx , by , bz}, a b = (ax i + ay j + az k ) (bx i + by j + bz k )= ax i (bx i + by j + bz k ) + ay j (bx i + by j + bz k ) + az k (bx i + by j + bz k )= ax bx i i + ax by i j + ax bz i k +ay bx j i + ay by j j + ay bz j k + az bx k i + az by k j + azbz k k = ax bx + ay by + az bz:a b = ax bx + ay by + az bz(1)
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:a ={ax, ay , az}, b = {bx , by , bz} ax bx + ay by + az bz = 0
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4. a ={ax, ay , az}, b = {bx , by , bz}, (1) a b .: Prjba = | a | |a | |b | = a b , (2)
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(2) a, b | a | | b |cos = a b, (3)
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2. M (1, 1, 1), A(2, 2, 1)B(2, 1, 2), AMB.:
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: F OM .:
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1. : ab, , c, c = ab(1) | c | = | a | | b | sin(2) c ab, ( c ac b). c a b .c a b , : a b.: c = a b: . ab, | a | | b |sin, a b, ab.
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2. (1) a a = 0 (2) a b = b a (3) a (b + c) = a b + a c (4) : (b + c) a = b a + c a ( a) b = a ( b) c = (a b)
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(5) a b a b = 0 : : a b, = 0 = . | a b | = | a | | b |sin = 0 a b = 0 : a b = 0 | a b | = | a | | b |sin = 0 | a | 0, | b | 0, = 0 = . a b
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: i i = j j = k k = 0 i j = k j k = i k i = j j i = k k j = i i k = j
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a ={ax, ay , az}, b = {bx , by , bz}, a b = (ax i + ay j + az k ) (bx i + by j + bz k )= ax i (bx i + by j + bz k ) + ay j (bx i + by j + bz k ) + az k (bx i + by j + bz k )= ax bx (i i) + axby ( i j ) + axbz( i k ) + ay bx (j i) + ayby ( j j ) + aybz (j k ) + azbx (k i) + azby ( k j ) + azbz( k k )= ax by k + ax bz( j) + ay bx(k) + ay bz i + az bx j + az by( i)= ( ay bz az by) i+( az bx ax bz) j+ ( ax by ay bx) k
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:a b =( aybz azby) i+( azbx axbz) j+ ( axby ay bx) k
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3: a = {2, 2, 1}b = {4, 5, 3}c.: a b ab= 6i + 4j + 10k 8k 6j 5i= i 2j + 2k c = a b = {1, 2 , 2}., 0R, c = {,2, 2} ab.
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4: ABCA(1, 2, 3), B(3, 4, 5), C(2, 4, 7), ABC.: .= 4i 6j + 2k
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.1. : SF (x, y, z) =0:(1) SF (x, y, z) =0;(2) SF (x, y, z) =0;, F (x, y, z) =0S, SF (x, y, z) =0.
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.1. M0(x0, y0, z0), R. : (x x0)2 + (y y0)2 + (z z0)2 = R2 (1)(1). M R: O(0, 0, 0), : x2 + y2 + z2 = R2 M(x, y, z), |M M0|2 =R2.
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: (x 1)2 + (y + 2)2 + z2 = 51: x2 + y2 + z2 2x + 4y = 0?
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2. :: x2 + y2 = R2.xoy, x2 + y2 = R2 O, R.xoy x2 + y2 = R2 . z L . z lxoyx2 + y2 = R2 , .
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(1) : C L .C. L .
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2: y2 =2x . z , xoyy2 =2x,.oxzyy2 =2x
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2: xy = 0. z , xoyxy = 0, z.xy = 0zxyo
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(2) .1 F (x, y) =0 : z , xoy C: F (x, y) = 0 .2 F (x, z) =0 : y , xoz C: F (x, z) = 0 .3 F (y, z) =0 : x , yoz C: F (y, z) = 0 .
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3. (1) : C, .
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: yozC, F (y, z) = 0, C z .M0(0, y0, z0)C, F( y0, z0) = 0C z M0M (x, y, z), z0 = z, F( y0, z0) = 0, :
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4: z = ay z .: y , :z2 = a2 ( x2 + y2 ), .
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5. xoz x z . : x z .
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1. : x, y, z:ax2 + by2 + cz2 +dxy + exz + fyz + gx + hy + iz +j = 0, . a, b, , i, j ..
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2 z = k( |k | c), |k | c , |k |, ; |k | = c , .2. .(1) 1 z = 0,
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3 , x = 0,y = 0, :: a=b=c, x2 + y2 + z2 = a2 , o, a.
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(2) : 1 z = k ,(k 0), z = k.k = 0, O(0,0,0); k, .2 y = k,
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3 x = k , .
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(3) ( p , q )
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(4)., (x z ::
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x , , (z ;: :
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(5) :
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6 x0 y0 ., .( x y )
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.
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S1, S2, :S1: F (x, y, z) = 0S2: G (x, y, z) = 0S1 , S2C,.C, C.(1)
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1: x 2 + y 2 + z 2 = 32 z = 2, x 2 + y 2 + z 2 = 32 z = 2
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: O, a. z .xOy, .
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Cx, y, zt.x = x (t)y = y (t) (2)z = z (t) t = t1, C(x, y, z), tC. (2).
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3: M x2 + y2 = a2 z , v z (,v), M , . : t, t = 0, x A(a, 0, 0), t, AM(x, y, z), MxOyM (x, y, 0).
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(1) z, t, AOM = t. x = |OM | cosAOM = acos ty = |OM | sinAOM = asin t(2) v z . z = MM = vt x = acos ty = asin tz = vt : .
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: = t. ; :x = acos y = asin z = b 0 0 + , zb 0 b 0+ b ,MOM ., = 2 , Mh = 2 b, h = 2 b.
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CF (x, y, z) = 0G (x, y, z) = 0(3)(3)zH (x, y) = 0 (4)(4)z , C .
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C, z (xOy)CxOy, xOyxOy, . CxOy.H (x, y) = 0z = 0: yOzxOz.
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4: :x2 + y2 + z2 = 1 x2 + (y 1)2 + (z1)2 = 1CxOy.: z ,z ,CxOy
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: z , x2 + y2 = 1z .C xoyx2 + y2 = 1z = 0xoy., xoy: x2 + y2 1
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1. :nn .: 1 , n;2 n .
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2. M0(x0, y0, z0), n={A,B, C}.:A(x x0) +B( y y0) +C( z z0) = 0(1) .(1)
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1: (2, 3, 0) n = {1, 2, 3}.: (1), :1 (x 2) 2 (y + 3) + 3 (z 0) = 0: x 2y + 3z 8 = 0
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: n.= 14i + 9j k2: M1(2, 1, 4), M2( 1, 3, 2)M3(0, 2, 3) ., :14(x 2) + 9(y + 1) (z 4) = 0: 14x + 9y z 15 = 0
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: A, B, C0, A 0, , n = {A, B, C}.: Ax + By + Cz + D = 0 (2).
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2: M0(1, 2, 3), 2x 3y + 4z 1= 0, .: n ={2 3, 4}2(x +1) 3(y 2) + 4(z 3) = 0: 2x 3y + 4z 4 = 0
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2. (1) O(0, 0, 0), D = 0. , :Ax + By + Cz = 0
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(2) xAx + By + Cz + D = 0, n = {A, B, C}x i ={1, 0, 0}, n i = A 1 + B 0 + C 0 = A = 0:x By + Cz + D = 0;y Ax + Cz + D = 0; z Ax + By + D = 0.: D = 0, .
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(3) xOy Cz + D = 0;xOz By + D = 0; yOz Ax + D = 0.
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3: x (4, 3, 1).: x , A = D = 0. By + Cz = 0(4, 3, 1), 3B C = 0 C = 3B By 3Bz = 0: y 3z = 0
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4: x, y, z P(a, 0, 0), Q(0, b, 0), R(0, 0, c), .: Ax + By + Cz + D = 0P(a, 0, 0), Q(0, b, 0), R(0, 0, c) , aA + D = 0bB + D = 0cC + D = 0:
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::(3)
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1. : ().:1: A1x + B1y + C1z + D1 = 0 n1 = {A1, B1, C1}2: A2x + B2y + C2z + D2 = 0 n2 = {A2, B2, C2}
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2. 12 A1A2 + B1B2 + C1C2 = 012 : 0, 0.
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6: M1(1, 1, 1)M2(0, 1, 1), x+y+z = 0, .: n ={A, B, C} x+y+z = 0 n1={1, 1, 1} :A (1) + B 0 + C (2) = 0 A 1 + B 1 + C 1 = 0
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: B=CA= 2CC = 1, n = {2, 1, 1}, 2 (x 1) + 1 (y 1) + 1 (z 1) = 0: 2x y z = 0
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7: P0(x0, y0, z0) Ax+By+Cz+D = 0, P0d.: P1(x1, y1, z1)P0P1NnP0 n ={A, B, C}:
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A(x0 x1) + B(y0 y1) + C(z0 z1) = Ax0 + By0 + Cz0 + D (Ax1 + By1 + C z1 + D) = Ax0 + By0 + Cz0 + D, P0Ax + By + Cz + D = 0:(4)
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: A(1, 2, 1): x + 2y + 2z 10 = 0
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12.1: A1x + B1y + C1z + D1 = 02: A2x + B2y + C2z + D2 = 0, L:A1x + B1y + C1z + D1 = 0A2x + B2y + C2z + D2 = 0L(1)(1)(1).
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1. : Ls ={m, n, p}, .s m, n, pL.sL
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2. LM0(x0, y0, z0) s ={m, n, p}(2).
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3. :x = x0 + mty = y0 + ntz = z0 + pt.(3)
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1: x + y + z +1 = 02x y + 3z + 4 = 0.: (1) M0(x0, y0, z0)z0 = 0, , x + y +1 = 02x y + 4 = 0:
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(2) s .1: x + y + z +1 = 0n1={1, 1, 1}2: 2x y+3z+4 = 0n2={2,1, 3}, = 4i j 3k, :
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2: A(2, 3, 4)B(4, 1, 3).: AB = {2, 2, 1},
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: , .L1, L2s1 ={m1, n1, p1}s2 ={m2, n2, p2}
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1. L1 L2:2. L1 L2 m1 m2 + n1 n2 + p1 p2 = 03. L1 L2
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: L1, L2 s1={1, 4, 1} s2={2, 2, 1}::
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:LL, L ., : s ={m, n, p} n ={A, B, C},
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(1) L : sin(2) L s // n ,(3) L s n,: Am + Bn + Cp = 0
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4: .: L s ={2, 7, 3} n ={4, 2, 2}s n = (2) 4 + (7) (2) + 3 (2) = 0M0(3, 4, 0)L, ,L , .
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: L s ={3, 2, 7} n ={6, 4, 14} L .
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: L s ={3, 1, 4} n ={1, 1, 1}s n = 3 1 + 1 1 + (4) 1 = 0L M0(2, 2, 3), , L .
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EMBED Equation.3 ,
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