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URGENT - C4 Vectors Question - +rep available watch

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    I can't do part (ii) of this question.

    I've worked out the equation easily enough, but I'm unsure of how to work out the coordinates of 'D'.

    I tried to equate x^2 + y^2 (one side from the vector equation I worked out, the other via the modulus of BD) but to no avail :dontknow:
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    The equation I have for BD is r = (-1, -7, 11) + lambda(15, -20, 0)

    That's correct according to the MS.
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    Bumping for halp :puppyeyes:
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    Right, you know the vector equation of the line BD and that its magnitude is 15.

    So work out the modulus of your vector equation for that line and set it equal to 15 so that you can find your lambda.

    You can then find D from this.
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    (Original post by marcusmerehay)
    Right, you know the vector equation of the line BD and that its magnitude is 15.

    So work out the modulus of your vector equation for that line and set it equal to 15 so that you can find your lambda.

    You can then find D from this.
    How do I work out the modulus of my vector equation (correctly)? When I do it I end up getting a constant... 661^0.5.

    Do I take the square root of the sum of the squared direction vector?
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    A vector equation should be of the form r = a + tb, where a and b are vectors in x,y,z and t is a variable.

    |r| = |a+tb|
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    Ohhhhhhhh.... :facepalm:

    Would I be right in thinking that |r| = t(625)^0.5 in this case? So that the modulus of a vector essentially ignores the 'a' bit of the r = a + tb, and is just the variable (t) * the modulus of the direction?
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    (Original post by Undulipodium)
    So if my equation is r = (-1, -7, 11) + t(15, 20, 0); and |r| = 15

    |r| = (1 + 225t^2 + 49 + 400t^2 + 121)^0.5 ???

    Sorry, it's late, my brain is melting. I'm a bit lost as to how to deal with the variable.
    Wait, i've just had a thought. You only need to worry about the 'tb' bit, so you require:

     |t(15,20,0)| = 15

     |t||(15,20,0)| = 15

     t = \frac{15}{|(15,20,0)|}

     t = \frac{15}{35} = \frac{3}{7}

    Or so I make it.
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    (Original post by marcusmerehay)
    Wait, i've just had a thought. You only need to worry about the 'tb' bit, so you require:

     |t(15,20,0)| = 15

     |t||(15,20,0)| = 15

     t = \frac{15}{|(15,20,0)|}

     t = \frac{15}{35} = \frac{3}{7}

    Or so I make it.
    So you just leave the variable and the position vector OUT of the calculation, and just work with the direction vector? I see where I was going wrong :p:

    I make it 3/5 (I think you've just typed 35 instead of 25 in the last line).

    Thanks very much :bigsmile:
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    (Original post by Undulipodium)
    So you just leave the variable and the position vector OUT of the calculation, and just work with the direction vector? I see where I was going wrong :p:

    I make it 3/5 (I think you've just typed 35 instead of 25 in the last line).

    Thanks very much :bigsmile:
    Indeed, my bad. :p:
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    (Original post by marcusmerehay)
    Indeed, my bad. :p:
    Thanks for all your help
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    Just out of interest which paper did that question come from please?
 
 
 
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Updated: January 15, 2010
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