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Vectors - vector equations

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    Okay, I'm a bit confused here, hoping someone can help clear this up.

    The question is "Points A, B and C have co-ords (0, 5), (9, 8) and (4, 3) respectively.
    a) Find the vector equation for the line joining A and B.
    b) Show that the perpendicular distance from C to line AB is 101/2 "

    I understand the vector equation = a + t(b - a), which works out for me as:
    (0, 5) + t(9, 3)
    But the answer at the back of the book says (0, 5) + t(3, 1)

    I'll be able to answer b) once I know how it works out at the above answer.
    Any help is appreciated
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    (Original post by Nagillum)
    Okay, I'm a bit confused here, hoping someone can help clear this up.

    The question is "Points A, B and C have co-ords (0, 5), (9, 8) and (4, 3) respectively.
    a) Find the vector equation for the line joining A and B.
    b) Show that the perpendicular distance from C to line AB is 101/2 "

    I understand the vector equation = a + t(b - a), which works out for me as:
    (0, 5) + t(9, 3)
    But the answer at the back of the book says (0, 5) + t(3, 1)

    I'll be able to answer b) once I know how it works out at the above answer.
    Any help is appreciated
     \begin{pmatrix} 9 \\ 3 \end{pmatrix} is the direction vector, it can be simplified as  \begin{pmatrix} 3 \\ 1 \end{pmatrix}

    It is similar to the gradient, if gradient  = \dfrac64 , this means moving 6 units in the x-direction and 4 units in the y-direction, this is similar to saying move 3 units in the x-direction and 2 units in the y-direction, i.e. Gradient  = \dfrac32

     \begin{pmatrix} 9 \\ 3 \end{pmatrix} = 3 \begin{pmatrix} 3 \\ 1 \end{pmatrix}
    Hence both vectors are parallel, any of them can be used.

    If you are still confused, then plot a point on a graph paper, and then use the vector to get another point on the line, you will see that both represent the same direction.

    Hope it makes sense.
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    (Original post by raheem94)
     \begin{pmatrix} 9 \\ 3 \end{pmatrix} is the direction vector, it can be simplified as  \begin{pmatrix} 3 \\ 1 \end{pmatrix}

    It is similar to the gradient, if gradient  = \dfrac64 , this means moving 6 units in the x-direction and 4 units in the y-direction, this is similar to saying move 3 units in the x-direction and 2 units in the y-direction, i.e. Gradient  = \dfrac32

     \begin{pmatrix} 9 \\ 3 \end{pmatrix} = 3 \begin{pmatrix} 3 \\ 1 \end{pmatrix}
    Hence both vectors are parallel, any of them can be used.

    If you are still confused, then plot a point on a graph paper, and then use the vector to get another point on the line, you will see that both represent the same direction.

    Hope it makes sense.
    I read the first sentence and went "OOHHHHHHHHHHHHHHHHH!" and now I feel stupid for not seeing that! Yeah I get it, thanks a lot man, 20 to midnight probably isn't the best time to be doing maths, eh :P
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    Your answer to the first part is the same as the one in the book, as the vectors (3,1) and (9,3) point in the same direction. [ (9,3) = 3*(3,1) ]
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    (Original post by RockEater)
    Your answer to the first part is the same as the one in the book, as the vectors (3,1) and (9,3) point in the same direction. [ (9,3) = 3*(3,1) ]
    Just to confirm something say I had the vector equation:
    (4, 12) + t(9, 3)

    I WOULD be able to cancel down the (9, 3) as it is a direction vector but I WOULD NOT be able to do this to the (4, 12) because it is a position vector...?
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    (Original post by Nagillum)
    Just to confirm something say I had the vector equation:
    (4, 12) + t(9, 3)

    I WOULD be able to cancel down the (9, 3) as it is a direction vector but I WOULD NOT be able to do this to the (4, 12) because it is a position vector...?
    Yes, you can simplify the direction vector but not the position vector.
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    Okay, cheers guys!

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