Finding vector $v_2$ given {$v_1,v_2$} is an orthogonal basis for set W

Question

I have a set $W=span${$x_1,x_2$}, where $x_1=(1,3,-3)$ and $x_2=(5,0,3)$. I am told that the set {$v_1,v_2$} is an orthogonal basis for the set $W$, and that $v_1=x_1=(1,3,-3)$, and I need to find $v_2$.

I started off by letting $v_2=(a,b,c)$. I need to find a system of three linear equations which I can solve for $a,b$ and $c$. Since the set is orthogonal, I know that the dot product of $v_1$ and $v_2$ is equal to zero which gives the first equation:

$$(1,3,-3)\cdot(a,b,c)=0$$ $$a+3b-3c=0$$

However I do not know what else I can do to find two other equations. Any help would be appreciated

Answer 1

$v_2$ is a linear combination of $x_1 $ and $x_2$. This gives a second linear equation. As any scalar multiple of $v_2$ also form a basis with $v_1$, we don't need a third equation.

Answer 2

The vector $v_2$ should be a linear combination of $x_1$ and $x_2$ so this amounts to solving a linear system $$c_1+5c_2=a\\ 3c_2=b\\-3c_1+3c_2=c$$ for $c_1$ and $c_2$. By reducing, you will find that the consistency of this linear system requires that $$3a-6b-5c=0.$$ You have already noted that $$a+3b-3c=0.$$ Now solve the above two for $a, b, c$ to get $$a=\frac{11}{5}c,\;\;\; b=\frac{4}{15}c$$ so that $v_2=c\begin{pmatrix} \frac{11}{5} \\ \frac{4}{15} \\ 1\end{pmatrix}$

You can also use Gram-Schmidt process if you are familiar with it.

Answer 3

Given $\vec{a} = (1,3,-3)$ , $\vec{b} = (5,0,3)$ .

Let a new orthogonal basis be $\vec{w}$ , $\vec{v}$.

Find an orthonal $\vec{w}$ , given that $\vec{v} = \vec{a}$.

Let $ \vec{w} = \alpha \vec{a} + \beta \vec{b} $.

Orthogonality:

$\vec{v} \cdot \vec{w} = 0$;

$\vec{a} \cdot (\alpha \vec{a} + \beta \vec{b}) = 0$.

$\alpha ||\vec{a}||^2 + \beta \vec{a} \cdot \vec{b} = 0$.

This is an equation for $\alpha, \beta$, more precisely for the ratio of $\alpha$ and $\beta$. Fixing the length of your orthogonal $\vec{w}$ will determine $ \alpha$ and $ \beta$.