证明:略. 5.在标准欧几里得空间内计算给下向量价内积,并求它的之间价夹角 (3)a=(3,-1,1,-1),B=( (4)a=(-1,1,-1,2,1),B=(3,1,-1,0,1) #f:(1)(a, B)=8,o, B)=arc cos (2)(a,B)=2,(a,B)=arco0 (3)(a,3)=-12,(a,B)= (4)(a,)=0.(a,)= 6.设x2+y2+2=1,(x,y,z∈R),试求 价最小值 解魔式=-3+1-x+1-y+1-2x,而魔标西布涅标夫斯基不等式 1-x2)2+(√1-y2)2+(√1-y v-2-ⅵ1-)(-z2)=3 明 2≥3, 所以 ≥-3+ 又当x=y=8=当时上式取等号因魔式价最小值为是 7.设a,b,c,x,,z∈R,列a2+b2+c2=25,x2+y2+2=36,am+by+cz=30.求旦+b+c价 值 解:魔标西-布涅标夫斯基不等式 30=ar +by +ca=(a b c) ≤√a2+b2+cm2+y2+2=30. 因等号成立时,(a,b,c)与(x,y,z)成相例设(a,b,c)=t(x,3,2),又入得 解而t 从而a+b+c: i. 5. kUS'Ppq{xg {, Ws89q5: (1) α = (1, 1, 1, 1), β = (−1, 2, 4, 3); (2) α = ³ 1 2 , −1, 1 3 , 1 6 ´ , β = (3, −1, 2, 2); (3) α = (3, −1, 1, −1), β = (−2, 2, −2, 2); (4) α = (−1, 1, −1, 2, 1), β = (3, 1, −1, 0, 1). : (1) (α, β) = 8, hα, βi = arc cos 2 √ 30 15 . (2) (α, β) = 7 2 , hα, βi = arc cos 7 10 . (3) (α, β) = −12, hα, βi = 5π 6 . (4) (α, β) = 0, hα, βi = π 2 . 6.  x 2 + y 2 + z 2 = 1, (x, y, z ∈ R), s x 2 1 − x 2 + y 2 1 − y 2 + z 2 1 − z 2 \. : K) = −3 + 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 . %NUV–WXUYdzUV), vuut µ 1 √ 1 − x 2 ¶2 + Ã 1 p 1 − y 2 !2 + µ 1 √ 1 − z 2 ¶2 · q ( p 1 − x 2) 2 + (p 1 − y 2) 2 + (p 1 − y 2) 2 > Ã 1 √ 1 − x 2 , 1 p 1 − y 2 , 1 √ 1 − z 2 !   √ 1 − x 2) p 1 − y 2 p 1 − y 2   = 3,  s 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 · √ 2 > 3, #$ 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 > 9 2 . x 2 1 − x 2 + y 2 1 − y 2 + z 2 1 − z 2 > −3 + 9 2 = 3 2 . Qb x = y = z = √ 3 3 Ry)zVY. !K)\" 3 2 . 7.  a, b, c, x, y, z ∈ R,  a 2 + b 2 + c 2 = 25, x 2 + y 2 + z 2 = 36, ax + by + cz = 30. s a + b + c x + y + z  . : NUV–WXUYdzUV), 30 = ax + by + cz = (a b c)   x y z   6 p a 2 + b 2 + c 2 · p x 2 + y 2 + z 2 = 30. !VY*+R, (a, b, c) B (x, y, z) *ej.  (a, b, c) = t(x, y, z), QRP 30 = t(x 2 + y 2 + z 2 ) = 36t, -% t = 5 6 . C% a + b + c x + y + z = 5 6 . · 8 ·