高等数学练习答案总习题七.docVIP

总习题七 1( 填空 (1)设在坐标系[O( i( j( k]中点A和点M的坐标依次为(x0( y0( z0)和(x( y( z)( 则在[A( i( j( k] 坐标系中( 点M的坐标为___________( 向量的坐标为___________( 解 M(x(x0( y(y0( z(z0)( ( 提示( 自由向量与起点无关( 它在某一向量上的投影不会因起点的位置的不同而改变( (2)设数(1、(2、(3不全为0( 使(1a((2b((3c(0( 则a、b、c三个向量是__________的( 解 共面( (3)设a((2( 1( 2)( b((4( (1( 10)( c(b((a( 且a(c( 则((____________( 解3( 提示( 因为a(c( 所以a(c(0( 又因为由a(c(a(b((a(a(2(4(1(((1)(2(10(((22(12(22)(27(9(( 所以((3( (4)设a、b、c都是单位向量( 且满足a(b(c(0( 则a(b(b(c(c(a(____________( 解 ( 提示( 因为a(b(c(0( 所以(a(b(c)((a(b(c)(0( 即 a(a(b(b(c(c(2a(b(2a(c(2c(a(0( 于是 ( (5)设|a|(3( |b|(4( |c|(5( 且满足a(b(c(0( 则|a(b(b(c(c(a|(____________( 解36( 提示( c(((a(b)( a(b(b(c(c(a(a(b(b((a(b)((a(b)(a(a(b(b(a(b(a(3a(b( |a(b(b(c(c(a|(3|a(b|(3|a|(|b|(3(3(4(36( 2( 在y 轴上求与点A(1( (3( 7)和点B(5( 7( (5)等距离的点( 解 设所求点为M(0( y( 0)( 则有 12((y(3)2(72(52((y(7)2(((5)2( 即 (y(3)2((y(7)2( 解得y(2( 所求的点为M(0( 2( 0)( 3( 已知(ABC的顶点为A(3,2,(1)、B(5,(4,7)和C((1,1,2)( 求从顶点C所引中线的长度( 解 线段AB的中点的坐标为( 所求中线的长度为 ( 4( 设(ABC的三边、、( 三边中点依次为D、E、F( 试用向量a、b、c表示、、( 并证明 ( 解 ( ( ( 5( 试用向量证明三角形两边中点的连线平行于第三边( 且其长度等于第三边长度的一半( 证明 设D( E分别为AB( AC的中点( 则有 ( ( 所以 ( 从而DE//BC( 且( 6( 设|a(b|(|a(b|( a((3( (5( 8)( b(((1( 1( z)( 求z ( 解a(b((2( (4( 8(z)( a(b((4( (6( 8(z)( 因为|a(b|(|a(b|( 所以 ( 解得z(1( 7( 设( |b|(1( ( 求向量a(b与a(b的夹角( 解 |a(b|2((a(b)((a(b)(|a|2(|b|2(2a(b(|a|2(|b|2(2|a|(|b|cos(a(^ b)( |a(b|2((a(b)((a(b)(|a|2(|b|2(2a(b(|a|2(|b|2(2|a|(|b|cos(a(^ b)( 设向量a(b与a(b的夹角为(( 则 ( ( 8( 设a(3b(7a(5b( a(4b(7a(2b( 求 ( 解 因为a(3b(7a(5b( a(4b(7a(2b( 所以 (a(3b)((7a(5b)(0( (a(4b)((7a(2b)(0( 即 7|a|2(16a(b(15|b|2 (0( 7|a|2(30a(b(8|b|2 (0( 又以上两式可得 ( 于是 ( ( 9( 设a((2( (1( (2)( b((1( 1( z)( 问z为何值时最小？并求出此最小值( 解 ( 因为当时( 为单调减函数( 求的最小值也就是求的最大值( 令( 得z((4( 当z((4时( ( 所以( 10( 设|