设空间两个单位向量向量OA=（m,n,0),向量OB=（0,n,p)与向量OC=(1,1,1）的夹角都等于π／4,

|OA|=1
=> m^2+n^2 = 1 (1)
OA.OC = |OA||OC|cos(π/4)
(m,n,0).(1,1,1) = 1. √3 ( √2/2)
m+n = √6/2 (2)
sub (2) into (1)
m^2 +(√6/2-m)^2 =1
4m^2 - 2√6m + 1 =0
m = (√6 +√2)/4 or (√6 -√2)/4
when m= (√6 +√2)/4 , n= (√6 -√2)/4
when m= (√6 -√2)/4 , n= (√6 +√2)/4

OA = ((√6 +√2)/4 , (√6 -√2)/4, 0) or ((√6 -√2)/4 ,(√6 +√2)/4,0)
Similarly,
OB.OC = |OB||OC|cos(π/4)
(0,n,p). (1,1,1) = 1. √3 (√2/2)
n+p = √6/2
p = √6/2 - n
when n = (√6 -√2)/4, p=(√6 +√2)/4
when n = (√6 +√2)/4, p=(√6 -√2)/4
OB =(0, (√6 -√2)/4, (√6 +√2)/4) or (0, (√6 +√2)/4, (√6 -√2)/4)