Our vectors are
#u_1=<-1,0,1>#
This vector #vecu_1# lies in the #x-z# plane and makes an angle of #pi/4# with the #z-#axis
#v_1=<-2,2,1>#
This vector #vecv_1# makes an angle of #arccos(2sqrt2/3)# with the #x-y# plane
Let the vectors we are looking for be #u_2# and #v_2#
Then, the unit vectors are
#hatu_2=<0,0,1>#
and
#hatv_2.vecv_1=<-2,2,1>.hatv_2#
#=||<-2,2,1>||*||hatv_2||*cos(pi/3)=3*1*1/2=3/2#
Let #hatv_2= < a,b,0 ># such that #sqrt(a^2+b^2)=1#, #=># #a^2+b^2=1#
and
#||<-2,2,1>|| * ||< a,b,0 >||*cos(pi/3)=3*1*1/2=3/2#
#-2a+2b+0=3/2#
#=>#, #a=(4b-3)/4#
Therefore,
#(4b-3)^2/16+b^2=1#
#16b^2-24b+9+16b^2=16#
#32b^2-24b-7=0#
#b=(24+-sqrt(24^2+4*32*7))/(2*32)=(24+-sqrt1472)/(64)=(24+-8sqrt23)/(64)=(3+-sqrt23)/(8)#
Therefore,
#b=(3+sqrt23)/8#, #=>#, #a=(((3+sqrt23)/2)-3)/4=(sqrt23-3)/8#
Their dot product is
#hatu_2.hatv_2= <0,0,1>.< a,b,0> =0+0+0=0#
Therefore,
#hatu_2# and #hatv_2# are perpendicular.
