Find 2 orthogonal vectors in a given subspace

Question

The problem itself is formulated as such:

Find 2 orthonormal vectors x,y so that span{x,y} = span{(0,1,1), (1,2,3)}

My understanding is:

span{(0,1,1), (1,2,3)} is a plane. Instead of finding 2 orthogonal vectors generating said subspace (plane), I can choose one of the vectors (f.e. (0,1,1)) and only need to find another vector from that plane, that is orthogonal to it (0,1,1). (Although I don't know how to do this)

I don't know if my understanding is correct, how could I solve this problem?

Answer 1

In general we have that given $v_1$ and $v_2$ we can construct $u_2$ orthogonal to $v_1$ by

$$u_2=v_2-\left(v_2 \cdot \frac{v_1}{|v_1|}\right) \frac{v_1}{|v_1|}$$

according to Gram–Schmidt process idea indeed

$$u_2 \cdot v_1=v_2\cdot v_1-\left(v_2 \cdot \frac{v_1}{|v_1|}\right) \frac{v_1\cdot v_1}{|v_1|}=v_2\cdot v_1-v_2\cdot v_1=0$$

Answer 2

Yes, your understanding is correct. Consider the vector$$\frac1{\sqrt2}(0,1,1)\left(=\frac{(0,1,1)}{\|(0,1,1)\|}\right).$$It belongs to your space and it is $(0,1,1)$ times a scalar. So,$$(1,2,3)-\left\langle(1,2,3),\frac1{\sqrt2}(0,1,1)\right\rangle\frac1{\sqrt2}(0,1,1)\tag1$$also belongs to your space. But $(1)$ is equal to $\left(1,-\frac12,\frac12\right)$, which is orthogonal to $(0,1,1)$. So, an answer to your problem is$$\left\{(0,1,1),\left(1,-\frac12,\frac12\right)\right\}.$$