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The vector equations of 2 beams of light r1 and r2 in 3 dimensional spaces referred to a fixed origin 0 are
r1=2i+3j+2k+^(5i-4j+3k)
r2=5i+5j-k+u(i-3j+3k)
where ^ and u are scalar parameters
a)prove that these lines intercept at a point in space
b)Find the position vector for the point at which these beams intercept
c)Find the acute angle between the beams

please help i am struggling with this question

  • maths -

    I don't like your use of ^ in this context.
    ^ is usually used to show exponents

    so
    r1 = (2i+3j+2k) + m(5i-4j+3k)
    or
    r1 = (2 + 5m , 3 - 4m , 2 + 3m)
    in the same way
    r2 = (5 +u , 5 - 3u , -1 + 3u)

    at their intersection:
    2+5m = 5+u and 3-4m = 5-3u
    5m - 3 = u and 3u = 2 + 4m
    sub the first into the second
    3(5m-3) = 2 + 4m
    15m - 4m = 2 + 9
    m = 1
    u = 2

    check for the 3rd component, that is, is 2+3m = -1 + 3u ?
    LS = 2+3m = 2+3=5
    RS = -1 + 3u = -1 +6 = 5

    yes, they intersect

    b) at what point?
    r1 = (2,3,2) + 1(5,-4,3) = (7,-1, 5)
    so the position vector to reach that point is
    r = 7i - j + 5k

    c)the direction vector of the first line is (5,-4,3)
    and of the second line is (1,-3,3)
    let Ø be the angle between them, you should know the definition of the dot product .....
    (5,-4,3)•(1,-3,3) = |(5,-4,3)| |(1,-3,3)| cosØ
    5 + 12 + 9 = √50√19 cos Ø
    cos Ø = 26/√950 = .....

    I got Ø = appr. 32.5° or .567 radians

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