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 $\vec{b}$ ?
Best Answer
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.