Answer
(a) Because the dot product ${\bf{v}}\cdot{\bf{w}}$ depends only on the angle between ${\bf{v}}$ and ${\bf{w}}$
(b)
${{\bf{e}}_1}' = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$
${{\bf{e}}_2}' = \left( {0,1} \right)$
${{\bf{e}}_1}\cdot{{\bf{e}}_2} = {{\bf{e}}_1}'\cdot{{\bf{e}}_2}'$
Work Step by Step
(a) Let $\alpha$ denote the angle between ${\bf{v}}$ and ${\bf{w}}$. By Theorem 2, the dot product of ${\bf{v}}$ and ${\bf{w}}$ is
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \alpha $
If both ${\bf{v}}$ and ${\bf{w}}$ are rotated by the same angle $\theta$, the angle between them is still $\alpha$. Therefore, the dot product ${\bf{v}}\cdot{\bf{w}}$ does not change.
(b) Write
${{\bf{e}}_1} = \left( {\cos {\theta _1},\sin {\theta _1}} \right) = \left( {1,0} \right)$,
${{\bf{e}}_2} = \left( {\cos {\theta _2},\sin {\theta _2}} \right) = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$,
where ${\theta _1} = 0$ and ${\theta _2} = \frac{\pi }{4}$.
So, the angle between ${{\bf{e}}_1}$ and ${{\bf{e}}_1}$ is $\frac{\pi }{4}$.
Now, rotating ${{\bf{e}}_1}$, ${{\bf{e}}_2}$ through an angle $\frac{\pi }{4}$ does not change the angle between them. This is the conclusion from part (a).
As results of this rotation, we get ${{\bf{e}}_1}'$ and ${{\bf{e}}_2}'$ :
${{\bf{e}}_1}' = \left( {\cos \left( {{\theta _1} + \frac{\pi }{4}} \right),\sin \left( {{\theta _1} + \frac{\pi }{4}} \right)} \right) = \left( {\cos \frac{\pi }{4},\sin \frac{\pi }{4}} \right)$
${{\bf{e}}_1}' = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$
and
${{\bf{e}}_2}' = \left( {\cos \left( {{\theta _2} + \frac{\pi }{4}} \right),\sin \left( {{\theta _2} + \frac{\pi }{4}} \right)} \right) = \left( {\cos \frac{\pi }{2},\sin \frac{\pi }{2}} \right)$
${{\bf{e}}_2}' = \left( {0,1} \right)$
Next, we verify that ${{\bf{e}}_1}\cdot{{\bf{e}}_2} = {{\bf{e}}_1}'\cdot{{\bf{e}}_2}'$.
${{\bf{e}}_1}\cdot{{\bf{e}}_2} = \left( {1,0} \right)\cdot\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 }}{2}$
${{\bf{e}}_1}'\cdot{{\bf{e}}_2}' = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)\cdot\left( {0,1} \right) = \frac{{\sqrt 2 }}{2}$
Hence, ${{\bf{e}}_1}\cdot{{\bf{e}}_2} = {{\bf{e}}_1}'\cdot{{\bf{e}}_2}'$.
