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    This seems to be slightly different than the rest of the vector questions, and it is slightly confusing.
    It says:

    Vectors r and s are given by

    r = ai + (2a-1)j - k
    s = (1-a)i + 3aj + (4a -1)k

    where a is a scalar.

    a) Find the values of a for which r and s are perpendicular.

    I don't really know how to go about doing this...for a start there are 2 positions in the book...and I don't see how these lines can cross more than once, unless the lines don't have to cross for them to be perpendicular to eachother...

    What I tried to do was to dot the directional vectors together and make that equivalent to 0 since this is when the lines would cross at 90*. But that wasn't the right asnwer...and also, how do I know that these two lines cross at one point with a 90* angle ( They don't seem to anyway)

    Thank you
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    This is what i would do:

    r.s = 0

    ( a ) . ( 1-a )
    ( 2a-1 ) . ( 2a )
    ( -1 ) . ( 4a-1)

    (a)*(1-a) + (2a-1)*(2a) + (-1)*(4a-1) = 0
    a - a^2 + 4a^2 - 2a - 4a + 1 = 0
    3a^2 - 5a + 1 = 0

    hmm does that factorise? Doubt it.

    { 5+/- SQURE ROOT [25-4*1*3] } / 6 =
    = Something quiet ugly = (5/6) +/- Square root [13/36]

    Now, if you plug that back into the top you'll get an answer but not very nice. Is that ANYWHERE near close? :/ I hate Vectors.
    • Thread Starter
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    Wait a sec, I am trying to work this out the answers are 1 and (1/5)
    EDIT argh I made a typo in the question...fixed now, but your working would now be:
    r.s = 0

    ( a ) . ( 1-a )
    ( 2a-1 ) . ( 3a )
    ( -1 ) . ( 4a-1)

    (a)*(1-a) + (2a-1)*(3a) + (-1)*(4a-1) = 0
    a - a^2 + 6a^2 - 3a - 4a + 1 = 0
    5a^2 - 6a + 1 = 0
    Which does factorise to give the correct values it's true but put these values back into the original equation... and these points don't even happen when the lines meet...
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    well if they are perpindicular the scalar product equals 0 ..
    the scalar product is i1*i2 + j1*j2 + k1*k2
    soo a(1-a) + 2a(2a-1) + -1(4a-1) = 0
    so we get a-a^2+4a^2-2a-4a+1 = 0
    3a^2-5a+1 = 0

    but there is no solution to this .. are u sure you gave me the number correctly?
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    thats what i did .. there seems to be sthg wrong ...
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    You know they cross each other at 90 degrees because the question says they are perpendicular to one another. I got the same as solly. What is the answer in the book?
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    r and s above are not equations of a line, but points (relative to the origin i presume. So you are looking for points that would be perpendicular from the origin.

    So just dot prouct r and s and solve to find a. you should get a quadratic that gives two solutions.

    I get, 5a^2 -6a+1=0
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    (Original post by silent ninja)
    r and s above are not equations of a line, but points (relative to the origin i presume. So you are looking for points that would be perpendicular from the origin.

    So just dot prouct r and s and solve to find a. you should get a quadratic that gives two solutions.

    I get, 5a^2 -6a+1=0
    how did u get that ??? :confused: i got 3a^2-5a+1 = 0
    refer to above
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    (Original post by Sollytear)
    This is what i would do:

    r.s = 0

    ( a ) . ( 1-a )
    ( 2a-1 ) . ( 2a )
    ( -1 ) . ( 4a-1)

    (a)*(1-a) + (2a-1)*(2a) + (-1)*(4a-1) = 0
    a - a^2 + 4a^2 - 2a - 4a + 1 = 0
    3a^2 - 5a + 1 = 0

    hmm does that factorise? Doubt it.

    { 5+/- SQURE ROOT [25-4*1*3] } / 6 =
    = Something quiet ugly = (5/6) +/- Square root [13/36]

    Now, if you plug that back into the top you'll get an answer but not very nice. Is that ANYWHERE near close? :/ I hate Vectors.
    should be 3
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    How did you get 5a^2?

    a(1-a) = a - a^2
    (2a-1)2a = 4a^2 - 2a
    -1(4a-1) = -4a +1

    add em gets you 3a^2 -5a + 1
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    o i just noticed the edit ..
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    (Original post by ma3eeni)
    well if they are perpindicular the scalar product equals 0 ..
    the scalar product is i1*i2 + j1*j2 + k1*k2
    soo a(1-a) + 2a(2a-1) + -1(4a-1) = 0
    so we get a-a^2+4a^2-2a-4a+1 = 0
    3a^2-5a+1 = 0

    but there is no solution to this .. are u sure you gave me the number correctly?
    that should be 3a
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    Indeed...
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    he just edited the question ^
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    (Original post by silent ninja)
    should be 3
    HeHe, a sneeky edit i see of the question

    Lets try it again then! moment while i type it all out
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    yep, i now get 1 or 1/5
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    okay you made me type this up now:

    a(1-a) + (2a-1)3a - (4a-1) = 0
    = a -a^2 + 6a^2 -3a -4a +1
    = 5a^2 - 6a + 1
    = (5a-1)(a-1)

    I didnt notice the op typed it up wrong, i read the Q from the book
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    This is what i would do:

    r.s = 0

    ( a ) . ( 1-a )
    ( 2a-1 ) . ( 2a )
    ( -1 ) . ( 4a-1)

    (a)*(1-a) + (2a-1)*(3a) + (-1)*(4a-1) = 0
    a - a^2 + 6a^2 - 3a - 4a + 1 = 0
    5a^2 - 6a + 1 = 0

    (5a-1)(a-1)
    a = 1/5 OR 1

    Add to start, and points are as follows:
    r = ai + (2a-1)j - k

    P] 1i + 1j - 1k
    Q] 1/5i - 3/5j - 1k

    PS: reason im doing this is so I know how to do it to
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    (Original post by Sollytear)
    HeHe, a sneeky edit i see of the question

    Lets try it again then! moment while i type it all out
    I read it from the book, its easier compared to the formatting of maths Q's on TSR.
    But your method was right
    • Thread Starter
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    Yeah umm, sorry for the confusion...
    A friends has helped me understand...basically I was treating r and s as equations of lines in vector form. Therefore, I couldn't get them to be perpendicular...I now understand that those two equations are actually points...
 
 
 
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