与a=(√2,√2)夹角为π/4的单位向量有2个：
(1,0)和(0,1)
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令所求向量b=(x,y),则：x^2+y2=1
b·a=(x,y)·(√2,√2)=√2x+√2y
故：cos=a·b/(|a|*|b|)=(√2x+√2y)/(2sqrt(x^2+y^2))=√2/2
即：sqrt(x^2+y^2)=x+y
即：xy=0,即：x=0或y=0
即：b=(1,0)或(0,1)