Does $a=\frac{\langle x,y\rangle}{\| y \|^2}$ imply $\| a \| =\frac{\| x \|}{\| y \|}$?

Question

As the title, um, $\overset{?}{\text{suggests}}$ I'm wondering if $a=\frac{\langle x,y\rangle}{\lVert y \rVert^2}\overset{\color{pink} \heartsuit}{\implies}\lvert a \rvert=\frac{\lVert x \rVert}{\lVert y \rVert}$. So what do you think?

EDIT$^1$:

Where $z=x-ay$ and $y$ and $z$ are orthogonal, with $y\neq 0$ and assuming that $$\lvert \langle x,y\rangle \rvert = \lVert x \rVert \cdot \lVert y \rVert.$$

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Perhaps this literature may be of help:

EDIT$^2$:

OK, I think I've got it know. Thank you for everyone's kind and considerate help.

Answer 1

No. If $x\perp y$, then $a=0$ even if $x\neq 0$.

If $\lvert\langle x,y\rangle\rvert=\lVert x\rVert \lVert y\rVert$ (that is, $x,y$ are colinear), then certainly $\lvert a\rvert=\frac{\lVert x\rVert}{\lVert y\rVert}$.

Not sure about the loving part, though.

Answer 2

The correct formula is $\dfrac{\langle x,y\rangle}{\|y\|^2}=\dfrac{\|x\|}{\|y\|}\cos\theta$, where $\theta$ is the angle between $x$ and $y$.