SOLUTION: find two vectors v1 and v2 whose sum is <3,-2> where v1 is parallel to <-4,5> while v2 is perpendicular to <-4,5> v1= v2=

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Question 1006183: find two vectors v1 and v2 whose sum is <3,-2> where v1 is parallel to <-4,5> while v2 is perpendicular to <-4,5>
v1=
v2=

Found 2 solutions by jim_thompson5910, AnlytcPhil:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Is there a typo? You mentioned the vector <-4,5> twice. Please double check. Thank you.
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
I don't understand why the other tutor thought it mattered that
< -4,5 > was mentioned twice.  Here is the picture of how it
is easily possible.  I'll put all the vectors with their tails 
at the origin:


Your problem was:

find two vectors v1 and v2 whose sum is <3,-2> where v1
is parallel to <-4,5> while v2 is perpendicular to <-4,5>.
As you can see from the graph, it certainly looks possible for v1
to be parallel to <-4,5> and v2 to be perpendicular to it.
And also the green lines show that the sum of v1 and v2 can
be <3,-2>. 

[I hope it doesn't bother you that v1 and <-4,5> in the graph are 
both on the same line and I said they are parallel.  Remember that 
vectors can be moved anywhere, and so vectors which are multiples 
of each other are considered to be parallel because they can be 
moved so that they are parallel by the usual standards of basic 
geometry.] 

So let's calculate what we see in the above drawing:

Since v1 is parallel to < -4,5 >, there is some non-zero scalar k 
such that

v1 = < -4k,5k >

let v2 = < a,b >

Since v2 is perpendicular to < -4,5 > its dot product 
with < -4,5 > is 0, so

v2•< -4,5 > = < a,b >•< -4,5 > = -4a+5b = 0

(1)    -4a+5b = 0

Since v1 + v2 = <3,-2> 

  < -4k,5k > + < a,b > = < 3, -2 >

(2)    -4k+a = 3
(3)     5k+b = -2

So we have this system of 3 equations in 3 unknowns:

  

The variable "a" is already eliminated in the 3rd equation,
so we eliminate "a" from the 1st and 2nd equations, by
multiplying the 2nd equation by 4 and adding it to the
1st equation:

-4a + 5b       =  0
 4a      + 16k = 12
-------------------
      5b + 16k = 12

Now we have the system of 2 equations in 2 unknowns:



Eliminate b by multiplying the second equation through 
by -5 and adding it to the first equation:

 5b - 16k = 12
-5b - 25k = 10
--------------
     -41k = 22
        k = 

Substituting in 

(2)    -4k+a = 3

We get:

      
      
      
      
      

Substituting in   

(3)     5k+b = -2

We get:

      
      
      
      
      

So the answer is:

v1 = < -4k,5k > =  = 

v2 = < a,b > = 

That's a pretty messy answer, but it's correct, and it looks
like it agrees with the drawing for v1 and v2 above. 

Edwin

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