anonymous
  • anonymous
For question 1A-4 a) Let P and Q be two points in space, and X the midpoint of the line segment PQ. Let O be an arbitrary fixed point; show that as vectors, OX = 1/2(OP + OQ). I am not sure how to follow the solution which is OX = OP + PX = OP + 1/2(PQ) = OP + 1/2(OQ−OP) = 1/2(OP + OQ).
MIT 18.02 Multivariable Calculus, Fall 2007
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SOLVED
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chestercat
  • chestercat
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IrishBoy123
  • IrishBoy123
|dw:1434488512478:dw| \(\vec {OX} = \vec {OP} + \vec {PX} \) \(= \vec {OP} + \frac{1}{2} \vec {PQ} \) \(= \vec {OP} + \frac{1}{2}( \vec {PO} + \vec {OQ}) \) \(= \vec {OP} + \frac{1}{2}( -\vec {OP} + \vec {OQ}) \) \(= \frac{1}{2} (\vec {OP} + \vec {OQ}) \)
IrishBoy123
  • IrishBoy123
sorry that drawing has really messed up, but the algebra is hopefully easy to follow
phi
  • phi
First, in general, identical vectors have identical length and direction *and it does not matter where their tail is located*. In other words, we can translate a vector and it does not change its identity. I assume you know that you can add vectors graphically by placing them "head to tail" thus adding OP to PQ (to get OQ) |dw:1434493583798:dw|

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phi
  • phi
we can write that as OP + PQ = OQ or if we subtract OP from both sides PQ= OQ - OP multiplying a vector by -1 flips the direction of the vector. now add head to tail you get a vector that represents the length and direction of PQ |dw:1434493938496:dw|
phi
  • phi
in the solution, hopefully by adding head to tail you see OX = OP + PX then by definition, PX is 1/2 the length (and same direction) as PQ, so OX= OP + 1/2 PQ and by the previous post PQ= OQ- OP OX = OP + 1/2 ( OQ - OP) distribute the 1/2 OX = OP +1/2 OQ - 1/2 OP combine OP - 1/2 OP to get 1/2 OP OX = 1/2 OP + 1/2 OQ factor out 1/2 OX = 1/2( OP+OQ)
phi
  • phi
If that is not clear, see https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/vectors/v/adding-vectors
anonymous
  • anonymous
Thank you both for the responses! It was silly not to think of it this way but glad you cleared it for me.

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