Answer
$\{(\frac{v_1}{||v_1||},\frac{v_2}{||v_2||})=\{(\frac{2}{\sqrt 30}+\frac{i}{\sqrt 30},-\frac{5i}{\sqrt 30},(-\frac{2}{\sqrt 6}-\frac{i}{\sqrt 6},-\frac{i}{\sqrt 6})\}$
Work Step by Step
We are given $v_1=(2+i,-5i)\\
v_2=(-2-i,-i)$
$v$ and $w$ are orthogonal if
$(v_1,v_2)=0 \\
= ((2+i,-5i),(-2-i,-i))\\
=(2+i)\bar {(-2-i)}+(-5i)\bar {(-i)}=0 \\
=(2+i)(-2+i)+(-5i)i \\
=0$
Hence, $v_1$ and $v_2$ are orthogonal vectors in $R^2$
To determine an orthogonal set, we obtain:
$||v_1||=\sqrt ((2+i,-5i)+(2+i,-5i))\\
=\sqrt (2+i)\bar {2+i}+(-5i)\bar {-5i} \\
=\sqrt (2+i)(2-i)+(-5i)5i\\
=\sqrt 30 $
$||v_2||=\sqrt ((-2-i,-i)+(-2-i,-i))\\
=\sqrt (-2-i)\bar {-2-i}+(-i)\bar {-i} \\
=\sqrt (-2-i)(-2+i)+(-i)i\\
=\sqrt 6 $
Hence, $\{(\frac{v_1}{||v_1||},\frac{v_2}{||v_2||})=\{(\frac{2}{\sqrt 30}+\frac{i}{\sqrt 30},-\frac{5i}{\sqrt 30},(-\frac{2}{\sqrt 6}-\frac{i}{\sqrt 6},-\frac{i}{\sqrt 6})\}$
