# Projection Of A Vector

Here is a neat little problem I have encountered:

What can be said about the vectors $\vec{u}$ and $\vec{v}$ if (a) the projection of $\vec{u}$ onto $\vec{v}$ equals $\vec{u}$ and (b) the projection of $\vec{u}$ onto $\vec{v}$ equals $0$?

For a), I believe the result stems from the fact that the two vectors are parallel and have the same magnitude.

For b), I know that the vectors would have to be orthogonal, but would they have to have the same magnitude?

I know i've answered these questions mostly on intuition, could someone help elaborate on my answers?

EDIT: $proj \large~\vec{u}_{~\vec{v}} = \frac{\vec{u} \cdot \vec{v}}{||\vec{v}||^2}\vec{v} \rightarrow \frac{||\vec{u}|| \cdot ||\vec{v}||\cos\theta}{||\vec{v}||^2}\vec{v} \rightarrow \frac{||\vec{u}||\cos\theta}{||\vec{v}||^2}\vec{v}$

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Start by writing down the formula for the projection. – Alex R. Feb 10 '13 at 16:54
In both cases you are right, except that the vectors do not need to have the same magnitude. – 1015 Feb 10 '13 at 16:58

The projection of $\vec{a}$ onto $\vec b$ is defined to be $$\mbox{proj}_{\vec{a}}\vec b = \frac {\vec{a} \cdot \vec{b}} {\left|\vec{b}\right|^2}{\vec{b}}.$$

So, for part (a), if $\mbox{proj}_{\vec{a}}\vec b = \vec{a}$, then $$\vec{a} = \frac {\vec{a} \cdot \vec{b}} {\left|\vec{b}\right|^2}{\vec{b}},$$ which only implies that $\vec b$ is parallel to $\vec a$. It says nothing about the magnitude of $\vec b$ because $$\left|\frac {\vec{a} \cdot \vec{b}} {\left|\vec{b}\right|^2}{\vec{b}}\right| = \left|\frac {\left|\vec{a}\right|\left|\vec{b}\right|\mbox{cos}\theta} {\left|\vec{b}\right|} \frac {\vec b} {\left|\vec{b}\right|}\right| = \left|\vec{a}\right|\mbox{cos}\theta,$$ where $\theta$ is the angle between the two vectors, and that formula doesn't depend on $|\vec b|$ at all.

For part (b), you know $\mbox{proj}_{\vec{a}}\vec{b} = \left|\vec{a}\right|\mbox{cos}\theta = 0$, so if $\left|\vec{a}\right| \ne 0$, then $\mbox{cos}\theta = 0$, so the vectors must be orthogonal. Again, it tells you nothing about their magnitudes.

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The magnitude of the projection of $\vec{u}$ onto $\vec{v}$ is

$$\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|} = |\vec{u}| |\cos{\theta}|$$

where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.

When $\vec{u}$ is parallel to $\vec{v}$, then $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}|$. When $\vec{u}$ is perpendicular to $\vec{v}$, then $\vec{u} \cdot \vec{v} = 0$.

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Hint: for (a): why the same magnitude? The projection of $(1,1)$ on $(2,2)$ equals $(1,1)$... (b) Again, why does the magnitude matter?

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