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IsTim

  • 4 years ago

Do the points A(2,1,5), b(-1,-1,10), and c(8,5,-5) define a plane? Explain why or why not.

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  1. phi
    • 4 years ago
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    use elimination to find if the three points are independent (not collinear)

  2. IsTim
    • 4 years ago
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    Elimination? -I think no, because there is no position vector or a direction vector.

  3. IsTim
    • 4 years ago
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    Pi?

  4. IsTim
    • 4 years ago
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    @Phi What is elimination?

  5. IsTim
    • 4 years ago
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    Sorry if I sound persistent, but I still don't understand (I have looked through the link).

  6. phi
    • 4 years ago
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    what kind of math are you studying?

  7. IsTim
    • 4 years ago
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    Vectors, of Vectors and Calculus.

  8. IsTim
    • 4 years ago
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    Specifically, Lines and Planes.

  9. phi
    • 4 years ago
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    Do they teach you about independent vectors? How to tell if two vectors are independent?

  10. IsTim
    • 4 years ago
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    I had difficulties understanding the course. I don't exactly know.

  11. IsTim
    • 4 years ago
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    equations of lines is the lesson (specifically)

  12. phi
    • 4 years ago
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    or maybe more simple: find the equation of a line through 2 of the points, and show the 3rd point does not satisfy the equation. So you know all 3 points do not lie on the same line, and therefore define a plane.

  13. IsTim
    • 4 years ago
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    What's an equation of a line?

  14. IsTim
    • 4 years ago
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    I'm just looking thru my textbook here, but I can only see equation of a plane. I'll use Google now though.

  15. IsTim
    • 4 years ago
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    y=mx+b? How does that work for a 3-space point?

  16. phi
    • 4 years ago
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    I think you do A + n(B-A) e.g. (2,1,5)+n ((-1,-1,10)-(2,1,5)) where n is any value (a scalar)

  17. IsTim
    • 4 years ago
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    You mean like a vector equation? [x,y,z]=[xo,y0,zo]+t[a1,a2,a3]+s[b1,b2,b3]

  18. IsTim
    • 4 years ago
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    But that wouldn't make sense to me...Aren't the ones with scalar multipliers direction vectors?

  19. phi
    • 4 years ago
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    B-A points in the direction from A to B |dw:1333072072947:dw| you scale it to move along it. Add A so the direction vector starts at A rather than the origin so A+ n(B-A)

  20. phi
    • 4 years ago
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    *should be B-A as the label

  21. IsTim
    • 4 years ago
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    Ok. Maybe I should ask what I should actually do for this question then. I don't seem to be able to comprehend our conversation.

  22. IsTim
    • 4 years ago
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    I just solve for that? All I had to do was plug them in at random onto a vector equation? Sorry, my eyes are starting to hurt.

  23. phi
    • 4 years ago
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    It looks like the 3 points are not collinear, so they define a plane

  24. IsTim
    • 4 years ago
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    Wait, how do I determine collinear? And, do I just do as what I suggested 2 comments above?

  25. phi
    • 4 years ago
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    See if this helps http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx

  26. IsTim
    • 4 years ago
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    How far should I read into this?

  27. phi
    • 4 years ago
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    Start at the paragraph below the ellipse.

  28. IsTim
    • 4 years ago
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    Ok. I just figured out that the equation of the line is the same as the vector equation.

  29. phi
    • 4 years ago
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    Here is the equation of the line through points (2,1,5) and (-1,-1,10) (2,1,5) + n( -1 -2, -1-1, 10-5) (2,1,5) +n( -3, -2, 5) Is there an n that gets us to the point (8,5,-5)? n= -2 gives us (2,1,5)-2(-3,-2,5) = (2+6, 1+4,5-10)= (8,5,-5) so it looks like there are all on the same line.

  30. IsTim
    • 4 years ago
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    ( -1 -2, -1-1, 10-5) I'm confused about that. I was never taught to do that in class. Also, "Is there an n that gets us to the point (8,5,-5)? n= -2 gives us (2,1,5)-2(-3,-2,5) = (2+6, 1+4,5-10)= (8,5,-5)" Do I figure out n's value by guess and check?

  31. phi
    • 4 years ago
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    No you don't guess. you have 3 separate equations for x,y,z example 2+n(-3) = 8 is the first one. It requires n= -2. it turns out all of the equations work for -2. so point (8,5,-5) is on the line

  32. phi
    • 4 years ago
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    that is point (-1,-1,10) - (2,1,5) rewritten as ( -1 -2, -1-1, 10-5)

  33. phi
    • 4 years ago
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    Look at Example 1 in Paul's notes

  34. IsTim
    • 4 years ago
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    I don't even understand that...

  35. IsTim
    • 4 years ago
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    IS he adding the 2 points or subtracting them?

  36. phi
    • 4 years ago
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    subtracting the 2 points. Maybe 2-d is easier? say you are at (1,1) and you want to get to (2,2) how much to you move in the x and y? subtract to find you have to move (1,1) the (1,1) represents your slope. to get from one point on the line to another move over 1 and up 1. Or scale the 1,1 to get to any point on the line. example you are at 1,1 and you move 1/2 over and 1/2 up. You are still on the line

  37. IsTim
    • 4 years ago
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    Ok. That makes sense I guess. I hope.

  38. phi
    • 4 years ago
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    To continue the 2-d example. if we say (x,y)= n(1,1) this means all points where y=x Say we want the line y= x+1 then we could say (x,y)= (0,1)+n(1,1)

  39. phi
    • 4 years ago
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    Good luck. At least you have the answer, It is not a plane.

  40. IsTim
    • 4 years ago
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    Ok. Thanks for the help, and putting up with my lack of knowledge on the topic.

  41. phi
    • 4 years ago
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    Does this help?

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