关于向量代数的几个小题（写出过程）

Question

1.设向量a的模=√3，b的模=1，a与b的夹角是π/6，求向量 a+b与a-b的夹角
2.设a+3b⊥7a-5b，a-4b⊥7a-2b，求a与b夹角
3.设向量a={2，-1，-2}，b={1，1，z}，问z为何值时a与b夹角最小，并求出此最小值
4.向量a的模=4，b的模=3，a与b的夹角是π/6，求以a+2b和a-3b为边的平行四边形面积

Answer 1

1.设 a+b与a-b的夹角为x
(a,b)=|a||b|cos(π/6)=3/2
(a+b,a-b)=|a|^2-|b|^2=2
(a+b,a+b)=|a|^2+|b|^2+2(a,b)=4+3=7, |a+b|=7^(1/2)
(a-b,a-b)=|a|^2+|b|^2-2(a,b)=4-3=7, |a-b|=1
cos(x)=(a+b,a-b)/[|a+b||a-b|]=2/[7^(1/2)]
x=arccos[2/7^(1/2)]
2.设a与b夹角为x.
0=(a+3b,7a-5b)=7(a,a)-15(b,b)+16(a,b)
0=(a-4b,7a-2b)=7(a,a)+8(b,b)-30(a,b)
0=[7(a,a)-15(b,b)+16(a,b)]-[7(a,a)+8(b,b)-30(a,b)]=-23(b,b)+46(a,b)
(b,b)=2(a,b)
0=7(a,a)-15(b,b)+16(a,b)=7(a,a)-30(a,b)+16(a,b)=7(a,a)-14(a,b)
(a,a)=2(a,b)
cos(x)=(a,b)/[(a,a)(b,b)]^(1/2)=(a,b)/[4(a,b)(a,b)]^(1/2)=1/2
x=π/3
3.设a与b夹角为x.
(a,b)=2-1-2z=1-2z
(a,a)=2^2+1+2^2=9,|a|=3.
(b,b)=1+1+z^2=2+z^2.
cos(x)=(a,b)/[|a||b|]=(1-2z)/[3(2+z^2)^(1/2)]
f(z)=(1-2z)^2/[9(2+z^2)]
f'(z)=(1/9)[2*2(2z-1)(2+z^2)-(1-2z)^2*2z]/(2+z^2)^2
=(2/9)(4+z)(2z-1)/(2+z^2)^2
lim_{z->负无穷}f(z)=4/9.
f(-4)=1/2>4/9.
f(1/2)=0.
lim_{z->正无穷}f(z)=4/9.
z<-4时f'(z)>0,f(z)单调增.4/9<f(z)<f(-4)=1/2.
-4<z<1/2时,f'(z)<0,f(z)单调减.1/2=f(-4)>f(z)>f(1/2)=0.
z>1/2时,f'(z)>0,f(z)单调增.4/9>f(z)>f(1/2)=0.
因此,f(z)在z=-4时取得最大值.f(-4)=1/2.
[cos(x)]^2=f(z)在z=-4处取得最大值.此时[cos(x)]^2=1/2.
z=-4时,夹角x=π/4最小.
4.设a+2b和a-3b的夹角为x.
面积=|(a+2b)|*|(a-3b)|*|sin(x)|
(a,b)=|a||b|cos(π/6)=4*3*3^(1/2)/2=2*3^(3/2).
(a+2b,a+2b)=|a|^2+4|b|^2+4(a,b)=16+36+4*2*3^(3/2)=4[13+2*3^(3/2)]
(a-3b,a-3b)=|a|^2+9|b|^2-6(a,b)=16+81-6*2*3^(3/2)=97-12*3^(3/2)
(a+2b,a-3b)=|a|^2-6|b|^2-(a,b)=16-54-2*3^(3/2)=-38-2*3^(3/2)
cos(x)=(a+2b,a-3b)/[(a+2b,a+2b)(a-3b,a-3b)]^(1/2)
[cos(x)]^2=(a+2b,a-3b)^2/[(a+2b,a+2b)(a-3b,a-3b)]
[sin(x)]^2 = 1 - [cos(x)]^2={(a+2b,a+2b)(a-3b,a-3b)-(a+2b,a-3b)^2}/[(a+2b,a+2b)(a-3b,a-3b)]
面积的平方=(a+2b,a+2b)(a-3b,a-3b)[sin(x)]^2
=(a+2b,a+2b)(a-3b,a-3b)-(a+2b,a-3b)^2
=4[13+2*3^(3/2)]*[97-12*3^(3/2)] -[-38-2*3^(3/2)]^2