Question #215421

Find vectors u; v € R^2 such that u is a scalar multiple of (1; 3), v is orthogonal to (1; 3); and (1; 2) =u+ v.



1
Expert's answer
2021-07-11T18:07:57-0400

From the question u=α(1,3)αR 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    α3y=1    y=α13also, 3α+y=2    y=23αequating the two values of y we have:23α=α13    69α=α1solving this we get α=710 from above we have that x=1α=1710=310also y=23α=23×710=110u=710(1,3)v=(310,110)\text{From the question } u=\alpha(1,3) \, \alpha \in \mathbb{R} \text{ also let }v=(x,y)\\ \text{then } v\cdot (1,3) =(x,y)\cdot(1,3)= x+3y =0 \implies x=-3y\\ \implies u+v=(\alpha+x,3\alpha+y)=(1,2) \\ \implies \alpha+x=1 \implies \alpha-3y=1 \implies y=\frac{\alpha-1}{3}\\ \text{also, }3\alpha +y=2\implies y=2-3\alpha\\ \text{equating the two values of $y$ we have:}\\ 2-3\alpha= \frac{\alpha -1}{3} \implies 6-9\alpha = \alpha -1 \,\, \text{solving this we get }\\ \alpha = \frac{7}{10} \text{ from above we have that } x=1-\alpha = 1- \frac {7}{10}= \frac {3}{10}\\ \text{also } y =2-3\alpha = 2-3\times \frac {7}{10} = \frac {-1}{10}\\ \therefore u=\frac {7}{10}(1,3)\,\, v=(\frac {3}{10},\frac {-1}{10})


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