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    http://filestore.aqa.org.uk/subjects...4-QP-JUN13.PDF question 6c. I've attached the diagram I constructed. My approach was to say that the point D would have co-ordinates (3-2λ,-2-3λ,4+2λ) according to the line worked out in b and then use the fact that the dot product of DC and DA must equal 0 since CDA is a right angle to find the value of lambda. Wrong lol.

    The mark scheme says that the dot product of CD and AD is 0 and goes from there. I don't get why because in the videos I watched the guy always said that whenever you're using the dot product, you have to take the vectors coming away from the angle, which according to my diagram would be DC and DA.

    Have I drawn my diagram wrong, misunderstood the dot product or got it completely wrong?
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    (Original post by Fudge2)
    ...
    It makes zero (haha, pun) difference.
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    (Original post by Zacken)
    It makes zero (haha, pun) difference.
    Maths jokes are always the best.

    Ok cool well that just means I made an error in adding some vectors which is infinitely more frustrating lol thanks
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    (Original post by Fudge2)
    Maths jokes are always the best.

    Ok cool well that just means I made an error in adding some vectors which is infinitely more frustrating lol thanks
    If you want to post a picture of your working, I'll be happy to take a look for you and point out what went wrong?
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    (Original post by Zacken)
    If you want to post a picture of your working, I'll be happy to take a look for you and point out what went wrong?
    Thanks a lot but I've spotted it according to me -4-(3-2λ) = -7-2λ. :facepalm:

    (apparently I can't rep you again yet, soz)
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    (Original post by Fudge2)
    Thanks a lot but I've spotted it according to me -4-(3-2λ) = -7-2λ. :facepalm:

    (apparently I can't rep you again yet, soz)
    Don't worry about the rep! Glad it's sorted now!
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    (Original post by Zacken)
    Don't worry about the rep! Glad it's sorted now!
    Another quick q actually if that's alright...when I work it through I get 17λ(λ+1)=0 so either λ=0 or λ=-1 (right answer). Am I right to say λ=0 can be discounted because like then the equation of the line wouldn't work kinda thing? Or I'd just be at the original like known point on the line? Not sure how to word it

    Edit 2 lol: do I say that λ=0 gives point A which obvs isn't what we're looking for?
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    (Original post by Fudge2)
    Another quick q actually if that's alright...when I work it through I get 17λ(λ+1)=0 so either λ=0 or λ=-1 (right answer). Am I right to say λ=0 can be discounted because like then the equation of the line wouldn't work kinda thing?
    I'm not entirely sure what you mean by the equation of the line wouldn't work?

    You can discount \lambda = 0 because that parameter returns the point A, i.e: we're saying D is A, which is nonsensical.
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    (Original post by Zacken)
    I'm not entirely sure what you mean by the equation of the line wouldn't work?

    You can discount \lambda = 0 because that parameter returns the point A, i.e: we're saying D is A, which is nonsensical.
    Haha cool I figured it out eventually (see edit). Thank you v much
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    (Original post by Fudge2)
    Haha cool I figured it out eventually (see edit). Thank you v much
    Yep, saw your edit. First class work!

    FWIW, in an exam, you wouldn't be required to justify the invalidity of \lambda (I think), so it's admirable that you're doing it now, it'll really build your understanding, nevertheless!
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    (Original post by Zacken)
    Yep, saw your edit. First class work!

    FWIW, in an exam, you wouldn't be required to justify the invalidity of \lambda (I think), so it's admirable that you're doing it now, it'll really build your understanding, nevertheless!
    Oh cool! Yeah I prefer to try to understand what's actually going on so I'll do it anyway
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    (Original post by Fudge2)
    Oh cool! Yeah I prefer to try to understand what's actually going on so I'll do it anyway
    I definitely agree that you should!
 
 
 
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