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    Been on this for a while really stumped, help would be appreciated.

    The line L1 has vector equation r = (2+\lambda)i + (-2-\lambda)j + (7+\lambda)k and line L2 has vector equation r = (4+4\mu)i + (26+14\mu)j + (-3-5\mu)k, where \lambda and \mu are scalar parameters.

    The vector n = -i +aj + bk where a & b are integers is perpendicular to both lines.

    Find the value of a and find the value of b.
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    If two vectors are perpendicular then their scalar product is 0. You can use this to derive two equations. However there are four unknowns here. Maybe you have to write a and b in terms of the two scalar parameters?
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    Oh dear my symbols got replaced with question marks, I hadn't realised. I used that LaTeX thing so it should make more sense now.
    So do I use the p.q = |p| \times |q| \times \cos \theta formula, making \theta equal to 90?
    And what do I put in for the values of p & q?
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    OK I don't think my last suggestion works at all. Your idea might have more promise. I'm guessing you'll have to apply that formula twice, once for L1 and r, and once for L2 and r. I assume you know how to work out scalar products and the modulus of a vector?
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    Yeah I already tried it with L1 and r and again for L2 and r but just ended up with a complete mess that I couldn't work anything out from so I'm not sure that's the way to go through with it.
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    where did u get this question from
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    (Original post by hazahaxa)
    Been on this for a while really stumped, help would be appreciated.

    The line L1 has vector equation r = (2+\lambda)i + (-2-\lambda)j + (7+\lambda)k and line L2 has vector equation r = (4+4\mu)i + (26+14\mu)j + (-3-5\mu)k, where \lambda and \mu are scalar parameters.

    The vector n = -i +aj + bk where a & b are integers is perpendicular to both lines.

    Find the value of a and find the value of b.
    Looking at L1 first of all. Just to make things clearer we'll rearrange into:

    r = (2i-2j+7k) +\lambda (i -j +k)

    Which is a standard form expressing the line as a postion vector and direction vector (in that order).

    When taking your dot product with n, you need to take it against the direction vector as that is what would be at right angles to n.

    so you'd have (i -j +k).(-i +aj + bk)= 0

    This will give you your first equation in a and b.

    Similarly for L2,....
 
 
 
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