Question

设a=(2,-1,-2),b=(1,1,z),问z为何值时<a,b>最小？并求出此最小值拜托各位了 3Q

Answer 1

解： |a| = 3,|b|=(2+z^2)^(1/2). a*b=2-1-2z=1-2z f(z)=a*b/[|a||b|]=(1-2z)/[3(2+z^2)^(1/2)] f'(z) = (1/3){-2(2+z^2)^(1/2)-(1-2z)z/(2+z^2)^(1/2)}/(2+z^2) = (1/3){-2(2+z^2) -(1-2z)z}/(2+z^2)^(3/2) = (1/3){-4-z}/(2+z^2)^(3/2) z < -4时，f'(z)>0,f(z)单调递增。 z > -4是，f'(z)<0,f(z)单调递减。 z = -4时，f(z)达到最大值f(-4)=(1+8)/[3(2+16)^(1/2)]=1/2^(1/2) 此时，夹角的最小值=arccos(1/2^(1/2))=PI/4.