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    Could anyone please explain the following question to me. Thanks

    Write down a vector equation of the straight line
    perpendicular to the xy-plane which passes through the point with coordinates (2, 1, 0)
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    (Original post by gooner1010)
    Can anyone please explain the following question to me. Thanks

    Write down a vector equation of the straight line
    perpendicular to the xy-plane which passes through the point with coordinates (2, 1, 0)
    Ok, let's break this down.

    For a vector équation of a straight line, we want to be able to start at any point which we know is on the line, and then follow the direction of the line to be able to get to any other point. That's the whole point of having an equation for it.

    Now, the direction of the line, you are told, is perpendicular (at a right angle to) to the x-y plane. So if you imagine the x-y plane drawn on a piece of paper, the line is actually coming straight up out of the paper. Can you express this direction as a vector?

    Now, the idea is to write the vector equation of the line (call it L) as

    L = a + \lambda b

    Where a is a point you know about, and b is the direction vector. We've multiplied this direction vector by the constant \lambda (which can be anything). So the idea is that you start at a, then choose how far you want to go in the direction that the line is going in. That choice is \lambda

    Hope this helps.
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    the perpendicular direction would actually be the z direction ? which has

    o
    o
    1

    as its vector. so this can be the normal n

    the vector equation of a plane where the normal is n is given by r.n = k

    since you know a point in the plane you can use its position vector as r, then find k by doing the dot product with n
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    (Original post by the bear)
    the perpendicular direction would actually be the z direction ? which has

    o
    o
    1

    as its vector. so this can be the normal n

    the vector equation of a plane where the normal is n is given by r.n = k

    since you know a point in the plane you can use its position vector as r, then find k by doing the dot product with n
    The OP's question asks for the vector equation of a line.

    Best to mention it, in case the OP got confused.
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    (Original post by Indeterminate)
    Ok, let's break this down.

    For a vector équation of a straight line, we want to be able to start at any point which we know is on the line, and then follow the direction of the line to be able to get to any other point. That's the whole point of having an equation for it.

    Now, the direction of the line, you are told, is perpendicular (at a right angle to) to the x-y plane. So if you imagine the x-y plane drawn on a piece of paper, the line is actually coming straight up out of the paper. Can you express this direction as a vector?

    Now, the idea is to write the vector equation of the line (call it L) as

    L = a + \lambda b

    Where a is a point you know about, and b is the direction vector. We've multiplied this direction vector by the constant \lambda (which can be anything). So the idea is that you start at a, then choose how far you want to go in the direction that the line is going in. That choice is \lambda

    Hope this helps.
    So what would be the answer?
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    (Original post by gooner1010)
    So what would be the answer?
    Well, it says "write down", so the idea is that you don't have to do much work.

    Can you identify the simplest vector that's perpendicular to the x-y plane?

    (someone's already pointed it out :lol:)
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    (Original post by notnek)
    The OP's question asks for the vector equation of a line.

    Best to mention it, in case the OP got confused.
    thank you for pointing that out....

    once you have the direction n

    0
    0
    1

    and the point on the plane with position vector a

    2
    1
    0


    you just write

    r = a + λn
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    (Original post by Indeterminate)
    Well, it says "write down", so the idea is that you don't have to do much work.

    Can you identify the simplest vector that's perpendicular to the x-y plane?
    (0,0,1), but how would i form the equation from this value?
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    (Original post by Indeterminate)
    Well, it says "write down", so the idea is that you don't have to do much work.

    Can you identify the simplest vector that's perpendicular to the x-y plane?

    (someone's already pointed it out :lol:)
    would it be (2,1,Sz) ?
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    (Original post by gooner1010)
    (0,0,1), but how would i form the equation from this value?
    Simple. It is the point you know + \lambda multiplied by the direction vector (which is what you've just written). \lambda is just a constant, remember. It can be anything of our choosing depending on how far we want to travel along the line.
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    if you don't like λ you can use µ instead.
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    (Original post by the bear)
    if you don't like λ you can use µ instead.
    But can he use \pi if he wishes?
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    (Original post by RDKGames)
    But can he use \pi if he wishes?
    Oh stop it :laugh:

    I hope no one uses e for this kind of thing either.
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    (Original post by Indeterminate)
    Oh stop it :laugh:

    I hope no one uses e for this kind of thing either.
    eeeeeeeeew
 
 
 
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