Question #222329

Construct an orthonormal basis for the subspace of R3 spanned by the vectors (1,-1,1)

and (2,0,4)


1
Expert's answer
2021-08-17T17:50:48-0400

Let us find aRa\in\R such that the vectors u=(1,1,1)\overline{u}=(1,-1,1) and v=(1,1,1)+a(2,0,4)=(1+2a,1,1+4a)\overline{v}=(1,-1,1)+a(2,0,4)=(1+2a,-1,1+4a) are orthogonal. Then the dot product of these vectors is equal to 0, that is 1+2a+1+1+4a=0.1+2a+1+1+4a=0. It follows that 6a=3,6a=-3, and hence a=12.a=-\frac{1}{2}.

Then v=(1+2(12),1,1+4(12))=(0,1,1).\overline{v}=(1+2(-\frac{1}{2}),-1,1+4(-\frac{1}{2}))=(0,-1,-1). Since u=1+1+1=3|\overline{u}|=\sqrt{1+1+1}=\sqrt{3} and u=0+1+1=2,|\overline{u}|=\sqrt{0+1+1}=\sqrt{2}, we conclude that the vectors uu=(13,13,13)\frac{\overline{u}}{|\overline{u}|}=(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}) and vv=(0,12,12)\frac{\overline{v}}{|\overline{v}|}=(0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) form an orthonormal basis for the subspace of R3\R^3 spanned by the vectors (1,1,1)(1,-1,1) and (2,0,4).(2,0,4).


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