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C4 - Vector parallel to the Z axis, why do I take it as (0,0,1)? Watch

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    Find a vector equation or the line which is parallel to the z-axis and passes through the point (4,-3,8).

    I know the third coordinate is the z value, and looking at the mark scheme, I can see that for a, you take a to be (0,0,1). Why is this though? Why do you know that the parallel to z axis bit makes the z value 1 and the x and y values 0?

    Also, for the equation r=a+tb, how do you know which is a and which is b in this case? The mark scheme says the (0,0,1) is the a value, but how do you know to take (0,0,1) to be a and not taking it to be b?
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    (Original post by vector12)
    Find a vector equation or the line which is parallel to the z-axis and passes through the point (4,-3,8).

    I know the third coordinate is the z value, and looking at the mark scheme, I can see that for a, you take a to be (0,0,1). Why is this though? Why do you know that the parallel to z axis bit makes the z value 1 and the x and y values 0?

    Also, for the equation r=a+tb, how do you know which is a and which is b in this case? The mark scheme says the (0,0,1) is the a value, but how do you know to take (0,0,1) to be a and not taking it to be b?
    Because you (should) know that the x-axis has direction 1 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}, then the y-axis has direction 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k}, and so the z-axis has direction 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}.

    These are all linearly independent vectors, which gives the perpendicular axes of the system. Hence, for a vector parallel to the z-axis, you can just pick (0,0,1), which is the same direction as the z-axis itself.

    The markscheme should say that a=(4,-3,8) and b=(0,0,1)
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    (Original post by vector12)
    Find a vector equation or the line which is parallel to the z-axis and passes through the point (4,-3,8).

    I know the third coordinate is the z value, and looking at the mark scheme, I can see that for a, you take a to be (0,0,1). Why is this though? Why do you know that the parallel to z axis bit makes the z value 1 and the x and y values 0?

    Also, for the equation r=a+tb, how do you know which is a and which is b in this case? The mark scheme says the (0,0,1) is the a value, but how do you know to take (0,0,1) to be a and not taking it to be b?
    Ok so remember the vector equation is in the form r=a+tb where is a particular position vector (in this case (4,-3,8)), t is a scalar constant and b is a directional vector. So essentially you need a line in the 3D plane that passes through (4,-3,8) so therefore this point must lie on the vector line. The easiest way to ensure this is let a be the position vector (4,-3,8) Note I really should be writing vectors with vertical columns but Idk how to .

    Now the other part of the equation is making the line parallel to the z axis- that must mean the directional vector parallel to the z axis must be in the form (0,0,m) where m is essentially any real number. The reason why the x and y values are 0 is to ensure the vector is parallel to thee z axis otherwise it wouldn't be- imagine the 3D coordinate axes and having any line parallel to the z axis and finding 2 points on that line - the two points will definitely have the same x and y values and the only thing that will be different will be there z values so if you imagine finding the vector that connects those two lines it will be (p-p, q-q, s-u) which means that x and y values become 0 and only the z value has a non zero number. The mark scheme has let m=1 for simplicity but technically it can be any real number as the scalar constant t can take any real value.

    therefore one possible vector equation for this line is r= (4,-3,8) + t(0,0,1)

    Hope this helped
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    (Original post by RDKGames)
    Because you (should) know that the x-axis has direction 1 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}, then the y-axis has direction 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k}, and so the z-axis has direction 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}.

    These are all linearly independent vectors, which gives the perpendicular axes of the system. Hence, for a vector parallel to the z-axis, you can just pick (0,0,1), which is the same direction as the z-axis itself.

    The markscheme should say that a=(4,-3,8) and b=(0,0,1)
    How do you know which is a and which is b though? And why should a=(4,-3,8) and not (0,0,1)?
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    (Original post by Anonymouspsych)
    Ok so remember the vector equation is in the form r=a+tb where is a particular position vector (in this case (4,-3,8)), t is a scalar constant and b is a directional vector. So essentially you need a line in the 3D plane that passes through (4,-3,8) so therefore this point must lie on the vector line. The easiest way to ensure this is let a be the position vector (4,-3,8) Note I really should be writing vectors with vertical columns but Idk how to .

    Now the other part of the equation is making the line parallel to the z axis- that must mean the directional vector parallel to the z axis must be in the form (0,0,m) where m is essentially any real number. The reason why the x and y values are 0 is to ensure the vector is parallel to thee z axis otherwise it wouldn't be- imagine the 3D coordinate axes and having any line parallel to the z axis and finding 2 points on that line - the two points will definitely have the same x and y values and the only thing that will be different will be there z values so if you imagine finding the vector that connects those two lines it will be (p-p, q-q, s-u) which means that x and y values become 0 and only the z value has a non zero number. The mark scheme has let m=1 for simplicity but technically it can be any real number as the scalar constant t can take any real value.

    therefore one possible vector equation for this line is r= (4,-3,8) + t(0,0,1)

    Hope this helped
    Thank you. How can you tell which is a position vector and which is a directional vector? And so how can you tell that a should be (4,-3,8) and not the (0,0,1)?
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    (Original post by vector12)
    How do you know which is a and which is b though? And why should a=(4,-3,8) and not (0,0,1)?
    It's basic vector equation fact. The a is some point that you want the line to go through, and b is the direction in which you want the line to go. Applying to your question, you get a=(4,-3,8) and b=(0,0,1)
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    the vector in the z direction does not have to be ( 0, 0, 1 )

    it could be ( 0, 0, 84 ) or

    ( 0, 0, -34.67 )

    but ( 0, 0, 1 ) keeps it nice & simple
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    (Original post by vector12)
    Thank you. How can you tell which is a position vector and which is a directional vector? And so how can you tell that a should be (4,-3,8) and not the (0,0,1)?
    Basically the position vector is always alone on its own so in this case it would be a. The directional vector is always multiplied with the scalar constant t so you know in this case it must be b. Think of a as simply representing a single point in a 3D space; b tells a which direction to move such that a can get to any other infinitely many points on the line
 
 
 
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