+0  
 
+2
234
2
avatar+7266 

Given   \(\vec{v}=3\vec{i}+4\vec{j}\)   and   \(\vec w=-2\vec i+7\vec j\)  ,  decompose  \(\vec v\)  into two vectors   \(\vec v_1\)   and   \(\vec v_2\)  , where   \(\vec v_1\)   is parallel to   \(\vec w\)   and   \(\vec v_2\)   is orthogonal to   \(\vec w\)   .

 

I need help understanding this...

I do know how to "get the answer" but I don't "get" the answer. smileylaugh

hectictar  Apr 25, 2018
 #1
avatar+88898 
+4

 

 

It's been awhile since I did this...so...here goes nothing

 

v = (3, 4)     w  =  (-2, 7)

 

We want to project  v onto w

 

projw v  =   [( v dot w) /( length of w)^2 ] * w  =  [ 22/ 53) * (-2, 7)  =  (-44/53, 154/53)

This  is   v1

 

To  find v2   we have    [ v - v1] =  (3, 4) - (-44/53, 154/53)  =  (203/53, 58/33)

 

Here's a pic :

 

 

The "why" of this always gave me some trouble, too. If I can remember what my  Calc teacher said, if we shine a beam onto v from the "top" of the diagram perpendicular to v2, it will "project" a perfect "shadow" of v onto v2. Likewise....if we shine a beam from the "right" of the diagram onto v such that the beam is perpedicular to v1, it will "project" a perfect "shadow" of v onto v1.

 

Sorry, hectictar.....that's about as good of an explanation as I can supply......!!!!

 

I think this is used in Physics to break up components of work  (as well as some other applications, too...)

 

cool cool cool

CPhill  Apr 25, 2018
 #2
avatar+7266 
+2

Oh okay thank you for that explanation. The picture does help.

 

HMMM I have been thinking about it for awhile now and I think I am starting to get it!!

hectictar  Apr 26, 2018

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.