# If the projection of vector $x$ onto vector $y$ is $y$, and projection of $y$ onto $x$ is $\frac12 x$, what is the angle between $x$ and $y$?

If the projection of vector $x$ onto vector $y$ is $y$, and projection of $y$ onto $x$ is $\frac12 x$, what is the angle between $x$ and $y$?

I calculated the angle between them to be $45^{\circ}$, and would like this answer to be verified.

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Well, how did you get this answer? (It's correct, by the way.) – TonyK Oct 18 '11 at 8:01

If $y$ is a side of a square, and $x$ the diagonal, then it will be true that the projections are as you say and the angle is as you say, so it looks good.

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$\triangle ABD \sim \triangle ABC \Rightarrow \frac{x}{2}:y=y:x$ ,(see picture bellow)

$y^2=\frac{x^2}{2}\Rightarrow y=\frac{x}{\sqrt{2}}\Rightarrow (BD)^2=y^2-\frac{x^2}{4}=\frac{x^2}{4}\Rightarrow BD=\frac{x}{2}$

If we apply sinus theorem we may write:

$$\frac{\frac{x}{2}}{\sin \alpha}=\frac{y}{\sin \frac{\pi}{2}}\Rightarrow \sin \alpha=\frac{\sqrt{2}}{2}\Rightarrow \alpha=\frac{\pi}{4}$$

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Without using a diagram, if we let $\theta$ be the angle between $\vec{x}$ and $\vec{y}$, we are given that $|x|\cos \theta=|y|,\ |y| \cos \theta =\frac{|x|}{2}$, so $(\cos \theta)^2=\frac{1}{2}$ and we need the positive solution to have the projections be positive. So $\cos \theta =\frac{\sqrt{2}}{2}, \theta=\frac{\pi}{4}$

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