Question

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

and (2,0,4)

Answer 1

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

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