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Math Help - Position Vectors

  1. #1
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    Position Vectors

    The question reads:

    With respect to the origin O, the points A,B,C,D have position vectors given by

    OA = 4i + k, OB = 5i - 2j - 2k, OC = i + j, OD = -i -4k

    (i) Calculate the acute angle between the lines AB and CD
    Completed this, answer is theta = 99.03degrees

    (ii) Prove that the lines AB and CD intersect

    (iii) The point P has position vector i + 5j + 6k. Show that the perpendicular distance from P to the line AB is equal to sqrt3
    (I can solve this one)
    ______________________________________________
    (ii) Prove that the lines AB and CD intersect

    Please could someone help me with a formula here?


    Thank you kindly
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  2. #2
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    Formula's for question (iii) as well would be appreciated.

    Thanks again
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  3. #3
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    Anyone?
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  4. #4
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    Ok I figured out (ii), could someone help me with (iii)?
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  5. #5
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    Quote Originally Posted by laoch View Post
    Ok I figured out (ii), could someone help me with (iii)?
    What you need (now and later) is somewhere in the attachment.
    Attached Files Attached Files
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  6. #6
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    Thank you


    I feel silly asking but could you explain this:
    Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
    where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.

    x,y,z would be your (x,y,z) vector right? Why does it say point, and where does u fit in? :S
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  7. #7
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    Quote Originally Posted by laoch View Post
    Thank you


    I feel silly asking but could you explain this:
    Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
    where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.

    x,y,z would be your (x,y,z) vector right? Why does it say point, and where does u fit in? :S
    I'm sorry but I don't understand your question.

    If you're referring to the first few lines of the attachment, the only reference to a point is where the general form of the parametric equations of the two lines is given.

    Are you familiar with finding the parametric equations of a line if you know a point on the line and a vector in the direction of the line?
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  8. #8
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    Not at all.

    I know a . b = 0 for a perpendicular line, but there are no parametric equations in my book :/


    When I work out a . b it doesn't equal 0, which is why I'm stuck lol. I think maybe I'm doing something wrong.

    I went through the pdf you posted, it all makes sense except for this:
    Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
    where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.
    Line 2: x = x'0 +v1s , y = y'0 +v2 s , z = z'0 +v3s ,
    where (x'0 , y'0 , z'0 ) is a point on line 2 and (v1, v2 , v3 ) is a vector in the direction of line 2.

    I dont understand what substitutions to make

    Thank you
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  9. #9
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    (i) Calculate the acute angle between the lines AB and CD
    Completed this, answer is theta = 99.03degrees
    : an acute angle is an angle which is inferior to 90 so I guess you have an error there.
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  10. #10
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    It works out right for the 2nd question though :S

    (i) OA + OB = AB = 9i -k - 2j

    (9, -2, -1)

    OC + OD = j - 4k
    (0, 1, 4)

    |AB| = sqrt86
    |CD| = sqrt17
    |a||b|cos theta = a1b1 + a2b2 + a3b3

    a.b = -6
    theta = -6/sqrt86sqrt17
    therefore theta = 99,03degrees



    *edit*

    -headdesks repeatedly-

    I was doing it the wrong way >.<

    New answer = 45,585degrees
    Last edited by laoch; September 23rd 2008 at 11:10 AM. Reason: mistake
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