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Math Help - Applied vector question

  1. #1
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    Question Applied vector question

    Hey there - I'm stumped with this piece of revision. Can't find many if any examples in my class notes. Can anyone offer some comments on my attempt and how I can complete this - Hopefully I can learn and use this as an example as ther are others.

    The position vectors of point A,B & C, relative to a fixed origin (O), are:
    a = 6i + 4j - k
    b = 8i + 5j - 3k
    c = 2i + 8j - 5k

    (i)Find the vector AB, the lenght AB, the cosine of angle ABC and the area of triangle ABC

    (ii) Find the vector equation of the straight line passing through A and B

    (iii) Find the co-ordss of the point D on the line for which OD is prependicular to AB. Hence, or otherwise calculate the shortest distance from O to the line AB

    OK, so for (i)
    AB = b - a
    AB = 2i + j - 2k

    Modulus of AB is 3

    To find the cosine of angle ABC I need the vector and modulus of CB,
    which is b -c = 6i - 3j +2k and mod. = 7

    The anlge is cos x = (AB * CB)/(mod. AB * mod. CB)

    cos x = 5/21

    For the area: The length/modulus of AC and CB multiplied and divided by 2

    mod. AC = 4*sqrt3
    mod. CB = 7

    Area therefore, 14sqrt3

    (ii) vector equation of line passing through A, B

    let P be any point on the line with vector r.

    OP = OA + AP

    AP = t*AB
    AP = t*(b-a)
    AB = (b - a) = 2i + j - 2k
    AP = 2ti + tj - 2tk

    therfore OP = OA + AP
    OP = 6i + 4j - k + 2ti + tj - 2tk
    OP = (6 +2t)i + (4+t)j - (1 + 2t)k

    (iii) no idea.

    How have I gotten on with my parts (i) & (ii)?

    Thanks, D
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  2. #2
    MHF Contributor
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    Hi

    I do not find the same as you for the area

    For (iii)
    D is on (AB) therefore from (ii) OD = (6 +2t)i + (4+t)j - (1 + 2t)k
    OD is perpendicular to AB therefore their dot product is equal to 0
    This gives you an equation to find t
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  3. #3
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    Yes, I've taken a look again - Not sure why I did that. area will be (mod.AB*mod.BC)/2

    So I get 21/2

    As for (iii) I calculate the coords as 2,2,3 and length sqrt17
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  4. #4
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    Wink

    Err, got it wrong again I see. I should use 1/2*AB*BC*Sin(arc cos(5/21)) which gives 10.198(3dp)
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  5. #5
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    Yes but using the cross product gives a simpler solution

    BTW
    sin²(arccos(5/21)) + cos²(arccos(5/21) = 1

    sin²(arccos(5/21)) + (5/21)² = 1

    sin²(arccos(5/21)) = 1 - (5/21)² = 416/21²

    sin(arcos(5/21)) = sqrt(416)/21 = 4sqrt(26)/21

    Area = 1/2 * 3 * 7 * 4sqrt(26)/21 = 2sqrt(26)

    You can easier find this solution using cross product
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