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    I'm not sure where to start..

    A = 2i+j+k
    B = 5i+7j+4k
    C = i - j

    Prove that the points A,B and C lie on a straight line l

    If all of the points are on the straight line that means their gradient is the same right? In order to find their grads i'd need a r = a + tb which I don't have..

    I'm really confused can someone help me?
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    (Original post by Jonario)
    I'm not sure where to start..

    A = 2i+j+k
    B = 5i+7j+4k
    C = i - j

    Prove that the points A,B and C lie on a straight line l

    If all of the points are on the straight line that means their gradient is the same right? In order to find their grads i'd need a r = a + tb which I don't have..

    I'm really confused can someone help me?
    The mathematical term for this is that they are collinear. So yes as you said, if AB and BC have the same direction vector they are parallel. To find the direction vector you don't need their equation for the vector line. You can work it out, as long as you know the position vecotrs of A, B and C. Also to show they are collinear, what can you say if AB and BC have the same direction vector, but also share the point B?
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    Lying on the same straight line is just another way of saying they are collinear. If their direction cosines are the same, then they are collinear. i.e A, B and C lie on a straight line.
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    (Original post by dknt)
    The mathematical term for this is that they are collinear. So yes as you said, if AB and BC have the same direction vector they are parallel. However, what can you say if AB and BC have the same direction vector, but also share the point B?
    If they both share the point B then there must be a connection somewhere?

    To answer this question would I have to find the direction vector for AB and BC ?
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    (Original post by forgottensecret)
    Lying on the same straight line is just another way of saying they are collinear. If their direction cosines are the same, then they are collinear. i.e A, B and C lie on a straight line.
    Do I have to find the angle between AB and BC ?
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    (Original post by Jonario)
    If they both share the point B then there must be a connection somewhere?

    To answer this question would I have to find the direction vector for AB and BC ?
    That's correct yes. If AB and BC are parallel, but also share point B, then they "connect" somewhere, but if they are also parallel, they must be collinear.

    And yes, you need to find their direction vectors i.e. \vec{AB} = \mathbf{b} -\mathbf{a} changing the vectors how you will.
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    (Original post by Jonario)
    If they both share the point B then there must be a connection somewhere?

    To answer this question would I have to find the direction vector for AB and BC ?
    Yes, that's literally a case of subtracting vectors i.e. \vec{AB} = \mathbf{b} - \mathbf{a}. etc.

    EDIT: dknt is too quick for me. :p:
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    (Original post by Farhan.Hanif93)
    Yes, that's literally a case of subtracting vectors i.e. \vec{AB} = \mathbf{b} - \mathbf{a}. etc.

    EDIT: dknt is too quick for me. :p:
    Sorry, I'm in a vector mood today!
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    (Original post by dknt)
    Sorry, I'm in a vector mood today!
    Ugh i'm using this C4 edexcel core book, my teacher decided not to follow it so my notes pretty much scrambled all over the place I'm still confused, I found out the direction vector for both AB and BC but they're not the same and i doubt this proves that they lie on l.. ?
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    (Original post by Jonario)
    Ugh i'm using this C4 edexcel core book, my teacher decided not to follow it so my notes pretty much scrambled all over the place I'm still confused, I found out the direction vector for both AB and BC but they're not the same and i doubt this proves that they lie on l.. ?
    They probably won't be the same. However, if  \vec{AB} // \vec{BC} then  \vec{AB} =  \lambda \vec{BC} In other words, one direction vector is a multiple of the other. (or should be at least if you've done it right)
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    (Original post by dknt)
    They probably won't be the same. However, if  \vec{AB} // \vec{BC} then  \vec{AB} =  \lambda \vec{BC} In other words, one direction vector is a multiple of the other. (or should be at least if you've done it right)
    :confused:

    What I got for AB is (3,6,3) and for BC is (-4,-8,-4)
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    (Original post by Jonario)
    :confused:

    What I got for AB is (3,6,3) and for BC is (-4,-8,-4)
    Try not to get vectors and co-ordinates mixed up. They're related, but not the same But you've done it right.

    (-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k}) = - \dfrac{4}{3}(3 \mathbf{i} + 6 \mathbf{j} + 3 \mathbf{i})

    hence, you've shown they are parallel. The only thing left is a conclusion, showing they are collinear.
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    (Original post by dknt)
    Try not to get vectors and co-ordinates mixed up. They're related, but not the same But you've done it right.

    (-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k}) = - \dfrac{4}{3}(3 \mathbf{i} + 6 \mathbf{j} + 3 \mathbf{i})

    hence, you've shown they are parallel. The only thing left is a conclusion, showing they are collinear.
    where did you get the -4/3 from ?
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    (Original post by Jonario)
    where did you get the -4/3 from ?
    If two vectors are parallel, one direction vector will be a multiple of the other. So, you need to find some scalar quantity to multiply to one direction vector to get it into the other, hence showing they are parallel.
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    (Original post by dknt)
    If two vectors are parallel, one direction vector will be a multiple of the other. So, you need to find some scalar quantity to multiply to one direction vector to get it into the other, hence showing they are parallel.
    Argh I'm still confused, I don't remember doing this. How to prove it step by step? Sorry about all the hassling, it's really frustrating when you spend 1hr trying to figure out how to do something.

    Btw the question says prove, arn't I mean't to show that A, B and C have something in common?
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    (Original post by Jonario)
    Argh I'm still confused, I don't remember doing this. How to prove it step by step? Sorry about all the hassling, it's really frustrating when you spend 1hr trying to figure out how to do something.
    You've basically done most the leg work.

    So you've gotten that \vec{AB}= (3 \mathbf{i} +6 \mathbf{j} + 3 \mathbf{k}) and that

    \vec{BC} = (-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k})

    You now want to show that they are parallel. If two vectors are parallel then,

    \vec{AB} = \lambda \vec{BC}, where lambda is a scalar quantity, so in other words when two vectors are parallel, their direction vectors will be a multiple of one another. If you can't spot what the scalar multiple is, divide one vector by the other (any way round).

    \dfrac{\vec{AB}}{\vec{BC}} : \dfrac{(3 \mathbf{i} +6 \mathbf{j} + 3 \mathbf{k})}{(-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k})}

    If I take out a factor of -\dfrac{4}{3} out of the numerator I'll get,


     \dfrac{-\frac{4}{3}(-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k})}{(-4 \mathbf{i} -8 \mathbf{j} -4 \mathbf{k})}

    The vectors top and bottom will cancel, leaving -\dfrac{4}{3}.

    Therefore, we can say that -\dfrac{4}{3} \vec{AB} = \vec{BC} or

     \vec{AB} = -\dfrac{3}{4} \vec{BC} and so therefore,

     \vec{AB} // \vec{BC} , however, if they are parallel, but share the point B they must be collinear.

    As you can see, the only thing you hadn't done is to show that they are parallel and therefore collinear.
 
 
 
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