From the question u=α(1,3)α∈R also let v=(x,y)then v⋅(1,3)=(x,y)⋅(1,3)=x+3y=0⟹x=−3y⟹u+v=(α+x,3α+y)=(1,2)⟹α+x=1⟹α−3y=1⟹y=3α−1also, 3α+y=2⟹y=2−3αequating the two values of y we have:2−3α=3α−1⟹6−9α=α−1solving this we get α=107 from above we have that x=1−α=1−107=103also y=2−3α=2−3×107=10−1∴u=107(1,3)v=(103,10−1)
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