Projection of one vector on another??

by Joza
Tags: projection, vector
Joza is offline
Dec3-07, 03:46 PM
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?
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Diffy is offline
Dec3-07, 03:59 PM
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does this site help?
EnumaElish is offline
Dec3-07, 11:20 PM
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projection of y onto x = x(x'x)-1x'y [= predicted value of y from the least squares equation "y = a + bx + u"].

HallsofIvy is offline
Dec4-07, 11:25 AM
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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 [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex], draw the line from the "tip" of [itex]\vec{u}[/itex] perpendicular with [itex]\vec{v}[/itex]. You now have a right triangle with angle [itex]\theta[/itex] between the angles and hypotenuse of length [itex]|\vec{u}|[/itex]. The length of the projection, the "near side", is then [itex]|\vec{u}|cos(\theta)[/itex]. Since the dot product can be defined as [itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex], to get the length of the pojection, we need to get rid of that [itex]|\vec{v}|[/itex] by dividing by it. The length of the projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex] is

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

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