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# P6 vectors watch

1. 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.
2. 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)
3. (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?
4. (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.
5. (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.
6. (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
7. (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.
8. Seeing as you might want to explore all avenues, finding the Cartesian equation of AB shouldn't be too hard.
9. 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.
10. [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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