⑩天掌 Teaching Plan on Advanced Mathematics 例1试用向量证明三角形的余弦定理 证设在△4BC中,∠BCA=BC=a,CA=b,AB=c 要证c2=a2+b2-2 ab cos6 记CB=a,CA=b,AB=c,则有 b =a-b 从而 B b c=cc=(a-b).(a-b)=a.a+bb-2a.b =la +6-2 alb cos(a, b) 由d=a,6=b,阳=c,及(Gb)=,即得 c-=a+b-2ab cos 0
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例1 试用向量证明三角形的余弦定理 记 CB a, = CA b, = AB c, = 则有 c a b, = − 从而 c c c a b a b a a b b a b = = ( − )( − ) = + − 2 2 2 cos( , ). 2 2 a b a b a b = + − 由 a = a, b = b, c = c, 及 (a,b) = , 即得 2 cos . 2 2 2 c = a + b − ab a b c A B C 设在 ABC 中， BCA = ,BC = a, CA = b, AB = c, 要证 2 cos . 2 2 2 c = a + b − ab 证：
⑩天掌 Teaching Plan on Advanced Mathematics 数量积的坐标表达式 ia=a i+a,j+a, k, b=b,i+b,j+b, k db=(ai +a,j+a, k). (b, i +b,j+b, k) i与jk,∵t·=jk=k·L=0, i闩jk=1, ii=jj=k·k=1. i·b=a.b.+a.b.+a.b x X
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 数量积的坐标表达式 a b = (a i a j a k) x y z + + (b i b j b k) x y z + + i j k, ⊥ ⊥ i j = j k = k i = 0, | i |=| j |=| k |= 1, i i = j j = k k = 1. x x y y z z a b = a b + a b + a b a a i a j a k, x y z = + + b bx i by j bzk 设 = + +
⑩天掌 Teaching Plan on Advanced Mathematics nb=l‖b|c0se→cs0=4:b ,(d≠0,b≠0 d‖b 两向量夹角余弦的坐标表示式: a.tab tab cos 6= 2 2 .+a.+a 6-+b+b 2 由此可知两向量垂直的充要条件为 ib<→a.b.+a,b+a.b.=0 ZZ
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics a b | a || b | cos = , | || | cos a b a b = 2 2 2 2 2 2 cos x y z x y z x x y y z z a a a b b b a b a b a b + + + + + + = 两向量夹角余弦的坐标表示式： a⊥b axbx + ayby + azbz = 0 由此可知两向量垂直的充要条件为 ( 0, 0). a b