Let u=(u1,u2),v=(v1,v2).
Given that u is a scalar multiple of (1,3)
u1=k,u2=3k Given that v is orthogonal to (1,3)
v1+3v2=0
Given that u+v=(1,2)
u1+v1=1u2+v2=2 We see that
u2=3u1v1=−3v2 Substitute
u1−3v2=13u1+v2=2 Then
v2=−3u1+2u1+9u1−6=1
u1=0.7v2=−0.1u2=2.1v1=0.3 u=(0.7,2.1)
v=(0.3,−0.1)
Comments