# Finding an orthonormal basis

#### fishingspree2

We have $$B=\left ( \overrightarrow{b_{1}},\overrightarrow{b_{2}} \right )$$ a basis of R^2 such as $$\left \| \overrightarrow{b_{1}} \right \|=1$$
$$\overrightarrow{b_{1}}\cdot \overrightarrow{b_{2}}=1$$ and $$\overrightarrow{b_{2}}=2$$. We also have $$\overrightarrow{e_{1}}=\overrightarrow{b_{1}}-\frac{1}{2}\overrightarrow{b_{2}}$$.
Find $$\overrightarrow{e_{2}}$$ such as $$C=(\overrightarrow{e_{1}},\overrightarrow{e_{2}})$$ is an orthonormal basis.

What I did:

We first construct $$B_{\perp }$$ by Gram-Schmidt.
$$\overrightarrow{b_{1\perp }}=\overrightarrow{b_{1}}$$
$$\overrightarrow{b_{2\perp }}=\overrightarrow{b_{2}}-\textup{proj}_{W_{1}}\overrightarrow{b_{2}}=\overrightarrow{b_{2}}-\frac{\overrightarrow{b_{2}}\cdot \overrightarrow{b_{1}}}{\overrightarrow{b_{1}}\cdot \overrightarrow{b_{1}}}\overrightarrow{b_{1}}=\overrightarrow{b_{2}}-\overrightarrow{b_{1}}$$

The magnitude of $$\overrightarrow{b_{2}}-\overrightarrow{b_{1}}$$ is sqrt(3). I got this using the law of cosines on a triangle with a Pi/3 angle and two sides of magnitude 1 and 2.

So $$B_{\perp }=\left (\overrightarrow{b_{1}},\frac{\overrightarrow{b_{2}}-\overrightarrow{b_{1}}}{\sqrt{3}} \right )$$.

$$\overrightarrow{e_{1}}$$ can then be written using transition matrix as $$\overrightarrow{e_{1}}=\frac{1}{2}\overrightarrow{b_{1}}-\frac{\sqrt{3}}{2}\left (\frac{\overrightarrow{b_{2}}-\overrightarrow{b_{1}}}{\sqrt{3}} \right )$$.

Let $$\overrightarrow{e_{2}} = x\overrightarrow{b_{1}}+y\overrightarrow{b_{2}} = \left (x+y \right )\overrightarrow{b_{1}}+\sqrt{3}y\left (\frac{\overrightarrow{b_{2}}-\overrightarrow{b_{1}}}{\sqrt{3}} \right )$$

Since $$B_{\perp }$$ is orthonormal and we want e2 and e1 orthonormal, we can write:
$$\frac{1}{2}\left ( x+y \right )-\frac{\sqrt{3}}{2}\sqrt{3}y=0 \Leftrightarrow x=2y$$

If y=1, then x=2

So $$\overrightarrow{e_{2}}=2\overrightarrow{b_{1}}+\overrightarrow{b_{2}}$$. By using law of cosines and a triangle with sides of 2 and 2, I get a magnitude of sqrt(12)

So $$\overrightarrow{e_{2}}=\frac{2\overrightarrow{b_{1}}+\overrightarrow{b_{2}}}{\sqrt{12}}$$.
Then $$C=(\overrightarrow{e_{1}},\overrightarrow{e_{2}})$$ is an orthonormal basis.

Any other, faster way to do it?

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