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I need to find a vector so as to extend basis with given vectors, $(-3,1,0)$ $(2,0,1)$ to $\mathbb{R}^3$.

I tried to orthogonalize the two using Gram Schmidt and then proceeded to find a third vector in the span of basis of Orthogonal Complement. I'm not sure if this is correct. Please help!

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2 Answers 2

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By using Gram Schmidt you get the vectors $\frac{1}{\sqrt{10}}(-3,1,0)$ and $\frac{1}{\sqrt{35}}(1,3,\frac{5 \sqrt{35}}{7})$. If you compute the dot product is zero. Now you can define a third vector (a,b,c) and impose the fact that $(a,b,c).\frac{1}{\sqrt{10}}(-3,1,0) =0$ and $(a,b,c).\frac{1}{\sqrt{35}}(1,3,\frac{5 \sqrt{35}}{7})=0$. You can determine a and b, while you're free to choose a value for c. You should get $(-\frac{1}{2},-\frac{3}{2},1)c$.

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  • $\begingroup$ I understand. Thanks for the answer. $\endgroup$
    – S.Rana
    Nov 12, 2018 at 1:21
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Ist approach:

Take the cross product of the given vectors. The resulting vector will be orthogonal to these two and the three of them will form a basis of $\Bbb{R}^3$.

2nd approach:

Find the span of the given vectors, you can determine that $$\text{Span} = \left\{\begin{bmatrix}x\\y\\z\end{bmatrix} \, \Big | x+3y-2z=0\right\}$$ Now choose a vector (for example, $\begin{bmatrix}1\\0\\0\end{bmatrix}$) which is NOT in this span. The three vectors will now form a basis.

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  • $\begingroup$ third approach: do it by inspection. $(0,-1,-1)$ $\endgroup$ Nov 11, 2018 at 19:37
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    $\begingroup$ @Matematleta :-) $\endgroup$
    – Anurag A
    Nov 11, 2018 at 19:38
  • $\begingroup$ This helps. Thanks. $\endgroup$
    – S.Rana
    Nov 12, 2018 at 1:21

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