Finding an orthonormal basis

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

What I did:

We first construct [tex]B_{\perp }[/tex] by Gram-Schmidt.
[tex]\overrightarrow{b_{1\perp }}=\overrightarrow{b_{1}}[/tex]
[tex]\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}}[/tex]

The magnitude of [tex]\overrightarrow{b_{2}}-\overrightarrow{b_{1}}[/tex] 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 [tex]B_{\perp }=\left (\overrightarrow{b_{1}},\frac{\overrightarrow{b_{2}}-\overrightarrow{b_{1}}}{\sqrt{3}} \right )[/tex].

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

Let [tex]\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 )[/tex]

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

If y=1, then x=2

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

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

Any other, faster way to do it?
 

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