Discussion in 'Homework Help' started by zerowingw, Sep 19, 2005.

  1. zerowingw

    Thread Starter New Member

    Sep 19, 2005
    Can anyone help me to solve this problem, becuase i can't solve this problem please help me.

    PQRS is a parallelogram and O is any point in the plane external to PQRS. If OP=p
    ,OR=r and OS=s, express the vector QO in terms of r, s and p.

    Thank you to take your take to answer this question for me and your time i please thank you so much.
  2. techduq

    New Member

    Aug 31, 2005

    Okay, let's agree on the basic orientation of the parallelogram: I'll let the top and bottom sides be parallel to the X-axis (not that it really matters), so the left and right sides can lean whichever way (both making the same angle wrt to the X-axis, of course). Furthermore, I'll identify the corners as follows: bottom left is P, then, moving clockwise around, we have Q, R, and S in the bottom right corner. I'll assume that when you say "OR" you mean the vector beginning at O and ending at R, etc.

    Now, we want a vector that runs from Q to O. So, if we start at point Q, let's try to find a way to get to P. Because once we are at P, all we have to do is add negative p (the vector) to get to O. Do you see that? Let v be the vector running from Q to P. Then the preliminary answer is v - p. See that? Well v is the vector from Q to P, but that's identical to the vector from R to S. And that is merely -r + s. Thus, V = s - r. Then vector QO = s - r - p. That's your answer, my friend.

    You may want to draw that out to follow the description easier.