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    2 submarines travelling in the ocean...

    r1=3i+4j-5k+£(i-2j+2k)
    r2=9i+j-2k+$(4i+j-k)

    Okay I found values for £ and $ earlier on when finding their point of intersection, do these values of £ and $ stay constant(ie i can put their vals in) when i try and answer the question "Point B has posn vector 10j-11k...show that only one of the submarines passes thru B" ?
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    yeah the £ and $ are just constants
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    i'm not sure u can actually keep the same vals tho...

    ARGHHH anyone know how to do the qu?
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    (Original post by little_wiggle)
    yeah the £ and $ are just constants
    just worked it out, and you;ve given bad advice matey, they're constants, however they can change if it that makes sense.

    Please be sure of what you're saying before you give advice next time.
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    yeah, they have to be able to change.....otherwise you would have r1 and r2 at the same positions every time
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    (Original post by BossLady)
    just worked it out, and you;ve given bad advice matey, they're constants, however they can change if it that makes sense.

    Please be sure of what you're saying before you give advice next time.
    there parameters not constants
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    you must remember that when dealing with vectors you are considering a line with a set length. Any line can be described by a direction and a pont it passes through. When dealing with this sort of vector the first part is the position vector of the point from the origin. The bit in brackets is the direction of the line and the coeficient of the bit in brackets tells you how far along the line away from the known point.

    MB
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    (Original post by musicboy)
    you must remember that when dealing with vectors you are considering a line with a set length. Any line can be described by a direction and a pont it passes through. When dealing with this sort of vector the first part is the position vector of the point from the origin. The bit in brackets is the direction of the line and the coeficient of the bit in brackets tells you how far along the line away from the known point.

    MB
    Exactly.

    Sub point B for r1 and r2
    and you can find that there is a real solution for £ but no for $
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    (Original post by BossLady)
    i'm not sure u can actually keep the same vals tho...

    ARGHHH anyone know how to do the qu?
    r1 = (3+λ)i + (4-2λ)j + (-5+2λ)k
    r2 = (9+4σ)i + (1+σ)j + (-2-σ)k

    part(i)

    at intersection,

    r1 = r2

    giving,

    λ=2
    σ=-1
    ===

    part(ii)

    if r1 passes thro' B, then

    (3+λ)i + (4-2λ)j + (-5+2λ)k = 0i + 10j - 11k

    3+λ=0
    4-2l=10
    -5+2l=11

    λ=-3 satisfies the above 3 eqns

    Therefore, r1 passes thro' B for λ=-3
    =======================

    if r2 passes thro' B, then

    (9+4σ)i + (1+σ)j + (-2-σ)k = 0i + 10j - 11k

    9+4σ=0
    1+σ=10
    -2-σ=-11

    from the 1st two eqns, we get

    σ=-2¼
    σ=9

    A contradiction,

    therefore r2 does not pass thro' B
    =====================
 
 
 

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