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    O is the origin, and A and B have position vectors a and b respectively relative to O, where a = 2i - j + 3k and b = -i + 3j -k

    Find a x b and hence write down the area of the triangle OAB

    find in the form of (r - u) x v = 0 an equation of the straight line AB.

    First bit noooo problem, second bit, guh?

    I guess it's along the lines of deducing if (r-u) and v are not zero length vectors, they must be parallel for (r - u) x v = 0. But then I'm a bit lost. Never seen this form before in my life. Help appreciated, rep available etc.
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    Did you get -8i - j + 5k for the first bit?

    Second bit, I'm not sure about all these different forms myself but as you say, if (r - u) x v = 0 the vectors (r - u) and v are parallel. If r is general posn vector of a point on the line, u is posn vector of any point on the line then (r - u) will be a vector parallel to the line so if (r - u) x v = 0 then v must also be a vector parallel to the line ie the direction vector (b - a)
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    (Original post by Bezza)
    Did you get -8i - j + 5k for the first bit?

    Second bit, I'm not sure about all these different forms myself but as you say, if (r - u) x v = 0 the vectors (r - u) and v are parallel. If r is general posn vector of a point on the line, u is posn vector of any point on the line then (r - u) will be a vector parallel to the line so if (r - u) x v = 0 then v must also be a vector parallel to the line ie the direction vector (b - a)
    Ah I can visualise that. how can you say that u is a position vector on the line though?
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    (r-u)xv = 0
    rxv-uxv = 0
    rxv = uxv

    I remember seeing a past paper asking to describe the locus of R in this case, the answer was a plane perpendicular to v.

    Hope that helps somehow.
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    (Original post by fishpaste)
    Ah I can visualise that. how can you say that u is a position vector on the line though?
    It seems the only way you'll get an equation describing that line. (r - u) is the vector joining the 2 points R and U, so v must be parallel to this vector for the cross product to be zero. If you had U anywhere else you'd be describing a different line.
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    (Original post by XTinaA)
    (r-u)xv = 0
    rxv-uxv = 0
    rxv = uxv

    I remember seeing a past paper asking to describe the locus of R in this case, the answer was a plane perpendicular to v.

    Hope that helps somehow.
    I think you're getting confused - the equation for a plane is r . n = d where r is posn vector on plane, n is normal vector to plane and d is a constant
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    (Original post by Bezza)
    I think you're getting confused - the equation for a plane is r . n = d where r is posn vector on plane, n is normal vector to plane and d is a constant
    Locus schmocus eh.
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    Seeing as you might want to explore all avenues, finding the Cartesian equation of AB shouldn't be too hard.
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    Just looked back at my past paper:

    (r-u)xv = 0, then the locus of r is a line through the point of u, parallel to v. So in this case, r is a line through A parallel to the vector OB. So instead of u and v, just write the position vectors of A and B respectively.
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    [r - (-i + 3j -k)] X (3i - 4j +4k) = 0 maybe.....

    Edit : ok... so shud have read all the other posts first...........tehehe
 
 
 

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