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    The line L1 is parallel to the vector (2i - j) and passes through the point with position vector (i+j-k). The line l2 passes through the points with position vectors (i + j - k) and (3i-2j+k). Find, to the nearest degree, the acuate angle between L2 and L2.

    Help please!
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    (Original post by imasillynarb)
    The line L1 is parallel to the vector (2i - j) and passes through the point with position vector (i+j-k). The line l2 passes through the points with position vectors (i + j - k) and (3i-2j+k). Find, to the nearest degree, the acuate angle between L2 and L2.

    Help please!
    If L1 si parallel to (2i - j), then it is going in that direction. Now find the equation of the line L2 (like you normally would), and use the scalar product with their direction vectors.
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    I get 40.6 = 41 degrees (to nearest degree)
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    (Original post by mockel)
    If L1 si parallel to (2i - j), then it is going in that direction. Now find the equation of the line L2 (like you normally would), and use the scalar product with their direction vectors.
    Ok what I did was:

    l1 = (i+j-k) + t(2i - J)

    L2 = (I+j-k) + x(2i - 3j - 2k)

    using scalar product

    (1x1) + (1x1) + (-1x-1) = root(2^2 + 1^2) x root(2^2 + 3^2 + 2^2) cos x
    3 = root5xroot17cosx
    3/root5xroot17 = cosx
    x= 71 degrees
    but the book gets 41, I cant see what Ive done wrong?
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    (Original post by imasillynarb)
    Ok what I did was:

    l1 = (i+j-k) + t(2i - J)

    L2 = (I+j-k) + x(2i - 3j - 2k)

    using scalar product

    (1x1) + (1x1) + (-1x-1) = root(2^2 + 1^2) x root(2^2 + 3^2 + 2^2) cos x
    3 = root5xroot17cosx
    3/root5xroot17 = cosx
    x= 71 degrees
    but the book gets 41, I cant see what Ive done wrong?
    Oh I get what Im doing wrong now :/ I need to multiply the direction vectors not the position vector thingies, I get 40.6 now...cheers!
 
 
 
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