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Linear Algebra Question

  1. Oct 5, 2014 #1
    ifx06a.jpg

    Please help me to understand the solution :( (I have made the sketch, is it correct?, where has the 1/4 comes from? is it like 1:3 (1 ratio 3) => numerator divided by numerator + denominator = 1/1+3 =1/4 (as given on page 6 at http://www.lowndes.k12.ga.us/view/14167.pdf [Broken] )? or am I wrong? For point position vector b in the solution the point B is on the left side from P (as A is on the right side of P)?)

    Thanks in advance.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 5, 2014 #2

    Mark44

    Staff: Mentor

    Your sketch isn't very accurate. It looks like A is more than 1/3 of the way along the segment between P and Q. From the problem description, the length of AQ is three times the length of AP, so A should be 1/4 of the way from P to Q.

    I don't understand what you're asking here. What point B?
     
    Last edited by a moderator: May 7, 2017
  4. Oct 5, 2014 #3

    Mentallic

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    It is 1/(1+3) = 1/4. If it were 1/3, then A would be a distance of x from P while A will be a distance of 2x from Q. Can you see this? This would make the ratio 1:2 as opposed to 1:3.

    Yes, except that the question doesn't introduce a new point B, it just asked whether A is uniquely determinable, which is no because it can be found on the left side of P as well. A has two possible positions.
     
    Last edited by a moderator: May 7, 2017
  5. Oct 6, 2014 #4

    Fredrik

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    The line through P and Q is the map ##t\mapsto p+t(q-p)##. Note that this map takes 0 to p and 1 to q. So if we restrict the domain of this map to [0,1], we get the line segment from P to Q. The expression p+t(q-p) can be interpreted as "start at p and take t steps of length |q-p| in the direction towards q". (I know that ##t\leq 1##, but you can certainly imagine taking less than one step; half a step would be a step that's half as long). How big a step would you have to take to end up at a point that's three times as far from P as from Q?

    The distinction between P and p is kind of nonsensical if you're working with the vector space ##\mathbb R^3##. It makes sense when you're working with a 3-dimensional affine space, but my guess is that this is from a book that doesn't mention affine spaces.

    For the second part, I would find all ##t\in\mathbb R## such that ##|a-p|=3|a-q|##, where ##a=p+t(q-p)=(1-t)p+tq##.
     
    Last edited: Oct 6, 2014
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