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    Q. Find a vector equation of the line that passes through the point (2,-1,6) and is perpendicular to the yz-plane.

    I know that the dot product of a.b has to equal 0, since cos 90 (ie meaning its perpendicular) is 0, but how can you work this out with only one set of points....?
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    (Original post by faa)
    Q. Find a vector equation of the line that passes through the point (2,-1,6) and is perpendicular to the yz-plane.

    I know that the dot product of a.b has to equal 0, since cos 90 (ie meaning its perpendicular) is 0, but how can you work this out with only one set of points....?
    Can you not just write r = (2 + a)i - j + 6k?

    Ben
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    (Original post by faa)
    Q. Find a vector equation of the line that passes through the point (2,-1,6) and is perpendicular to the yz-plane.

    I know that the dot product of a.b has to equal 0, since cos 90 (ie meaning its perpendicular) is 0, but how can you work this out with only one set of points....?
    If it's perpendicular to the yz plane then the yz values will not change, as the line runs through it. An easier way to think about it is a line which is perpendicular to the x axis, it goes through the x axis once and the x coordinate does not change, only y.

    Hence, another point on the line (the choice is arbitrary i think), would be one with the same yz values, but any x value for example (a,-1,6), where a is any number.

    From there you have two points, so you can form a vector equation of the line.
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    the answer in the back of the book is r=2i-j+6k+ lamda i

    so apart from the lamda i, isnt all three coordinates exactly the same- instead of 2???
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    (Original post by Ben.S.)
    Can you not just write r = (2 + a)i - j + 6k?

    Ben
    Sorry, I should explain my reasoning;

    If the line if perpendicular to the y-z plane, it can only point directly through that plane - along the x-axis. So the line is fixed at -1 on the y and 6 on the z axes and the x component (i) alone may vary. To find the vector equation of the line, get onto the line using a position vector from the origin (the first part of the equation) and then express the direction the line is travelling in as a scalar (I have used 'a') multiplying a direction vector.

    To get to the line, follow the vector 2i - j + 6k
    To travel along the line any distance use the vector of arbitrary length ai

    Putting these two together you get r = 2i - j + 6k + ai
    Factorising, you get r = (2 + a)i - j + 6k.

    Ben
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    jesus that was so simple- i try to picture things in my head to make them seem much clearer- it was just working out which axis it was perpendicular to that i had probs with- cheers
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    (Original post by faa)
    jesus that was so simple- i try to picture things in my head to make them seem much clearer- it was just working out which axis it was perpendicular to that i had probs with- cheers
    Are you a girl, by any chance?

    Ben
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    I'm a guy and i find it hard to picture 3 dimensions in my head. Thus, some of the obscure vector questions are the most difficult part of P3.
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    hang on- he quoted my "jesus" and then you answered for me!!! lol im a guy, but yeh- i prefer algebra, vectors shudnt be in p3, shud just stick to mechanics lol
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    (Original post by faa)
    hang on- he quoted my "jesus" and then you answered for me!!! lol im a guy, but yeh- i prefer algebra, vectors shudnt be in p3, shud just stick to mechanics lol
    Oh well - must just be myth. Just wait until vector calculus!

    Ben
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    I know he was referring to you, i just thought i'd mention that i find it hard to envision it too, you aren't alone.
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    vector calculus- which unit is that in?

    Well im not sure if you need to know P1 and 2 for the P3 vectors, but i got put in the class that have already done that (i am also in the p1,2 class aswell before you panick), but he just told me to have a go at that and see how i go, im not doing my p3 and 4 until jan
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    You don't need any prior A-level knowledge for P3 vectors. It's a stand alone chapter. As is P6 matricies and P4 complex numbers.

    (Don't quote me on any of that...)

    Vector calculus is degree level.

    It is nice to use a 3D graphing program and see exactly what the answers look like.
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    i also believe p4 series and sequences is - because aslong as u know the formulae you are ok :-)

    Well im taking p1,2 s1,d1, m1 in nov, so ive worked out if i can do a chapter a day (ie one chapter= p1 algebra) then i'll have 2 weeks free before the exams.........
 
 
 
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