# Projection of one vector on another??

by Joza
Tags: projection, vector
 P: 139 Can anyone explain how to find the projection of one vector along another? I thought it was scalar (dot) product, but then I realised it WASN'T. What is this then? Anyone explain?
 P: 443 does this site help? http://www.math.oregonstate.edu/home...d/dotprod.html
 Sci Advisor HW Helper P: 2,483 projection of y onto x = x(x'x)-1x'y [= predicted value of y from the least squares equation "y = a + bx + u"].
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## Projection of one vector on another??

The LENGTH of the projection of one vector onto another is (almost) the dot product.

To find the projection of $\vec{u}$ on $\vec{v}$, draw the line from the "tip" of $\vec{u}$ perpendicular with $\vec{v}$. You now have a right triangle with angle $\theta$ between the angles and hypotenuse of length $|\vec{u}|$. The length of the projection, the "near side", is then $|\vec{u}|cos(\theta)$. Since the dot product can be defined as $\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$, to get the length of the pojection, we need to get rid of that $|\vec{v}|$ by dividing by it. The length of the projection of $\vec{u}$ on $\vec{v}$ is
$$\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}$$

In order to get the projection vector itself, we need to multiply that length by the unit vector in the direction of $\vec{v}$, which is, of course, $\vec{v}/|\vec{v}|$.
The vector projection of $\vec{u}$ on $\vec{v}$ is
$$\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}$$

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