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anonymous

  • 5 years ago

Find an equation of the following lines. The line through (0,0, 1) parallel to the y-axis. I don't get why the answer is r(t)= <0, 0, 1> + t<0, 1, 0>

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  1. anonymous
    • 5 years ago
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    this is parametric form of the line

  2. amistre64
    • 5 years ago
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    parallel to omits the component its paralelling right?

  3. amistre64
    • 5 years ago
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    we can pick to point; or vectors that point to 2 points and create a line from them P(x,y,z) = (0,0,1) another point that is parallel to the y mean that the ys are the same right? for instance (0,0,0) and (0,6,0) i guess i had that backewards lol

  4. amistre64
    • 5 years ago
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    same others, diff y then lol

  5. amistre64
    • 5 years ago
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    (0,0,1) and lets say (0,4,1) should be parallel to the y right?

  6. amistre64
    • 5 years ago
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    it draws out right then :) we need a vector from one point to the other now; <0,4,1> works good right?

  7. anonymous
    • 5 years ago
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    so do you mean that i can pick any point as long as y value stay the same and z value is 1?

  8. anonymous
    • 5 years ago
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    m not getting u at all :(

  9. amistre64
    • 5 years ago
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    well, the unit vector; tha thas the same direction but a magnitude of 1 would be: <0,1,1>

  10. amistre64
    • 5 years ago
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    nah; the x and z are the same; the y can be any value

  11. amistre64
    • 5 years ago
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    P(0,0,1) and Q(0,1,1) are in line and parallel to the y axis

  12. amistre64
    • 5 years ago
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  13. anonymous
    • 5 years ago
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    can the answer be r(t) = <0, 0, 1> + t<0, any y value here, 1>?

  14. amistre64
    • 5 years ago
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    this is the view looking straight at the x axis; and y is to the left and right and z i up and down.

  15. amistre64
    • 5 years ago
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    the only way a line can be parallel to the y axis is to have z and x remain constant between points

  16. amistre64
    • 5 years ago
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    yes it can; but to make things simple; they use a "unit" vector

  17. amistre64
    • 5 years ago
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    so its length is 1; and then any value fo y just makes it scalar

  18. amistre64
    • 5 years ago
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    the line is a point plus a vector right? r(t) = P(0,0,1) + t<0,1,1> where t is a scalar amount

  19. amistre64
    • 5 years ago
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    i missed that 0 at the end :)

  20. amistre64
    • 5 years ago
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    ahh i see it; the vector is saying tha tit doesnt change from x to z; just the y value lol

  21. amistre64
    • 5 years ago
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    you see that?

  22. amistre64
    • 5 years ago
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    <0,1,0> means that it never stears away from the original x and z values; it just heads down y

  23. anonymous
    • 5 years ago
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    oh i see. I see it now. but where did they get the vector <0, 1, 0>?

  24. amistre64
    • 5 years ago
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    any vector pointing in the same direction will work; so they simply choose what is called a "UNIT" vector. It has the same direction; but its length is equal to "1"

  25. amistre64
    • 5 years ago
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    its just a standard for notation really

  26. amistre64
    • 5 years ago
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    its the direction you want ; the length of the vector doesnt matter

  27. amistre64
    • 5 years ago
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    as long as it aint a zero vecotr that is lol

  28. anonymous
    • 5 years ago
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    just to make sure I totally understand this concept. So if i write r(t) = <0,0,1 > + t<0, 4, 0>, i'm still correct right?

  29. amistre64
    • 5 years ago
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    yes :)

  30. anonymous
    • 5 years ago
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    thank you so much to help me understand it :)

  31. amistre64
    • 5 years ago
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    thank you for letting me :) the more i help the more I learn myself lol

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