
#1
Dec307, 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? 



#2
Dec307, 03:59 PM

P: 443

does this site help?
http://www.math.oregonstate.edu/home...d/dotprod.html 



#3
Dec307, 11:20 PM

Sci Advisor
HW Helper
P: 2,483

projection of y onto x = x(x'x)^{1}x'y [= predicted value of y from the least squares equation "y = a + bx + u"].




#4
Dec407, 11:25 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

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 [tex]\frac{\vec{u}\cdot\vec{v}}{\vec{v}}[/tex] 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 [tex]\frac{\vec{u}\cdot\vec{v}}{\vec{v}^2}\vec{v}[/tex] 


Register to reply 
Related Discussions  
Vector projection problem  Linear & Abstract Algebra  2  
Fundamental vector projection question  Calculus & Beyond Homework  1  
Vector calculus: Projection of a point to a plane  Calculus & Beyond Homework  2  
Vector Projection Proof  Introductory Physics Homework  6 