We know that $\vec{i}$ and $\vec{j}$ are unit vectors along $x$-axis and $y$-axis respectively. So angle between them is $90°$.
But let $\vec{a}$ and $\vec{b}$ be unit vectors such that $\vec{a}-\sqrt{2}\vec{b}$ is also a unit vector? How do we determine the angle between $\vec{a}$ and $\vec{b}$ ?
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$\begingroup$If $a - \sqrt{2} b$ is a unit vector, then the inner product \begin{equation*} \langle a - \sqrt{2}b, a - \sqrt{2}b \rangle = ||a - \sqrt{2} b||^{2} = 1. \end{equation*}
My suggestion would be to work out the inner product, first using bilinearity, and then using the relationship between the inner product of two vectors, their magnitudes, and their angles.
$\endgroup$ $\begingroup$The vector $a-\sqrt{2}b$ is unit so $|a-\sqrt{2}b|^2=1$ that is $$1=(a-\sqrt{2}b).(a-\sqrt{2}b)=a.a-2\sqrt{2}a.b+2b.b=2-2\sqrt{2}a.b$$ this concludes that $a.b=\dfrac{\sqrt{2}}{4}$. The angle between $a$ and $b$ is $$\cos\theta=\frac{a.b}{|a||b|}=\frac{a.b}{|a||b|}=\dfrac{\sqrt{2}}{4}$$ finally $$\color{blue}{\theta=\arccos\dfrac{\sqrt{2}}{4}}$$
$\endgroup$ $\begingroup$The vector $a-\sqrt{2}b$ is unit vector so $|a-\sqrt{2}b|^2=1$ therefore $$1=(a-\sqrt{2}b).(a-\sqrt{2}b)=a.a-2\sqrt{2}a.b+2b.b=3-2\sqrt{2}a.b$$ Implies that $a.b=\dfrac{2}{2\sqrt{2}}$. So angle between $a$ and $b$ is $$\cos\theta=\frac{a.b}{|a||b|}=\frac{a.b}{|a||b|}=\dfrac{1}{\sqrt{2}}$$ finally $$\color{green}{\theta=\arccos\dfrac{1}{\sqrt{2}} = π/4}$$
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