Vector Geometry

  • Thread starter twofish
  • Start date
  • #1
twofish
25
0

Homework Statement


Below is the question given.
"Give a vector description of a point P that is on-third of the way from A to B on the line segment AB."

2. The attempt at a solution
I'm not sure where to go with this, all I really have at this point is a line A-B 6 squares long on my graph paper, with the point P at square 2.
From there I have drawn a two vectors above the line, they both originate at O and descend down to A and B, such that another vector descends from O to P at 90 degrees.

I believe that I have to use the Proj of u onto v at some point (or b onto a), but I'm unsure how to incorporate this into the solution.

Code:
          O
         /|\
        / |  \
  a    /  |    \    b
      /   |       \
     /    | p       \
    /     |           \
   A______|_____________B
          P

 
Last edited:

Answers and Replies

  • #2
berkeman
Mentor
63,669
14,806
If you have the (x,y) coordinates of A and of P, then just do the component subtractions to get the vector AP.
 
  • #3
twofish
25
0
Yes, the question is as stated (ie; there are no components given).

I'm leaning towards something similar to letting
AB= b-a
then P= (1/3) (b-a)
?
 
Last edited:
  • #4
berkeman
Mentor
63,669
14,806
Then I guess you can only answer it with symbollic vector subtraction.

Assume that you have the vectors OB and OA. Write the vector subtraction that results in the vector AB. Then what modification to this subtraction do you need to do to represent the vector AP?
 

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