# Extending a Basis to R3

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!

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 $$(-\frac{1}{2},-\frac{3}{2},1)c$$.

• I understand. Thanks for the answer. Nov 12, 2018 at 1:21

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.

• third approach: do it by inspection. $(0,-1,-1)$ Nov 11, 2018 at 19:37
• @Matematleta :-) Nov 11, 2018 at 19:38
• This helps. Thanks. Nov 12, 2018 at 1:21