Determining a positive orthonomal basis

Question

So, there was this question that appeared on my linear algebra exam some hours ago, and i don't really understand how to solve:

Consider the vectors $\vec{u}$ = 2$\vec{i}$+$\vec{j}$+$\vec{k}$ and $\vec{v}$ = $\vec{i}$+2$\vec{j}$.

a) Determine a positive orthornomal basis {$\vec{a}$,$\vec{b}$,$\vec{c}$} with $\vec{a}$ parallel to $\vec{u}$ and $\vec{b}$ coplanar with $\vec{u}$ and $\vec{v}$.

b) Determine the coordinates of $\vec{w}$ = 3$\vec{i}$+4$\vec{j}$+5$\vec{k}$ in the orthonormal basis {$\vec{a}$,$\vec{b}$,$\vec{c}$}.

So, that's it. I think know how to find $\vec{a}$... since it is parallel to $\vec{u}$ i would just need to normalize $\vec{u}$ ($\vec{a}$ would be ${\vec{u}\over|\vec{u}|}$). But i'm stuck in how i would find $\vec{b}$ and $\vec{c}$. Hope y'all can help me.

Answer 1

For $b$ to be coplanar with $u, v$ then it is in the span of $u, v$. Y

You replaced $u$ with $\tilde{u} = u/\|u\|$. You can just take $v$, make it orthogonal to $\tilde{u}$, then normalize.

By "make orthogonal to", I mean take $\tilde{v} = v - \langle v, \tilde{u}\rangle v$ in the span of $u$ and $v$ so that $\langle \tilde{v}, u \rangle = 0$. This is just Gram-Schmidt.

Answer 2

Since you’re working in $\mathbb R^3$, you can use cross products to generate a basis with the requisite properties: $\vec u\times\vec v$ is orthogonal to both, while $(\vec u\times\vec v)\times\vec u$ is orthogonal to $\vec u$ and $\vec u\times\vec v$, i.e., it lies in the plane spanned by $\vec u$ and $\vec v$. Normalize and order these vectors so that the basis has the desired orientation.