已知向量a,b,c,a+c=b,a⊥c,且|a|=|c|= √2 （1）求|b|（2）求向量b与c所成的夹角θ的大小
(1) |b|^2=b^2=(a+c)^2=a^2+2a*c+c^2
=|a|^2+0+|c|^2
=4
|b|=2
(2) cos=b*c/|b|*|c|
=(c*(a+c))/2√2
=c^2/2√2
=2/2√2
=√2/2
=45°