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1. I know the answer to the first part of the question, but I don't know why, and I don't get the second part. Can anyone help?

A variable point P has position vector r

a) For a non-zero constnat vector c, describe the locus of point P in the case:

c cross product r = 0

b) In the case that P lies on the line with Cartesian Equations

(x-1)/2 = (y+1)/3 = (z-5)/4

find the equation of the line in the form

(r-a) cross product b = 0
2. a) The line through O parallel to c.

b) The equation (r - a) x b = 0 represents the line through a parallel to b. The points (1, -1, 5) and (3, 2, 9) are on the line (x-1)/2 = (y+1)/3 = (z-5)/4. Therefore (3, 2, 9) - (1, -1, 5) = (2, 3, 4) is parallel to that line. So we can take

a = (1, -1, 5),
b = (2, 3, 4).
3. (Original post by Mat666)
I know the answer to the first part of the question, but I don't know why, and I don't get the second part. Can anyone help?

A variable point P has position vector r

a) For a non-zero constnat vector c, describe the locus of point P in the case:

c cross product r = 0

b) In the case that P lies on the line with Cartesian Equations

(x-1)/2 = (y+1)/3 = (z-5)/4

find the equation of the line in the form

(r-a) cross product b = 0
For part, a), if you take the cross product, the magnitude of it is found by:
|r||c|sin T

Thus, if the cross product is 0 and c is non-zero, then either r is 0 (in which case the locus is just the origin), or sin T=0, in which case the locus is a line through the origin in the direction of c (which also includes the point in the origin), so the solution is a line through the origin in the direction of c.
4. Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
5. (Original post by Mat666)
Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
Consider the line L in 2d that goes through the points (0, 1) and (1, 0). By drawing a picture you can see that L is parallel to the vector (1, 0) - (0, 1) = (1, -1). It's the same in 3d (but more difficult to draw).
6. (Original post by Mat666)
Thanks: just a little clearing up. I understand how you got the points on the line, but why take one away from the other? I'm being really stupid today and its too hot, I usually get P6 vectors...
All you need is the position vector of a point on the line (a) and a point on the line (b). You can find a by finding any point that satisfies the equation of the line. To get b, all you need is two such points that satisfy the equation of the line then find the vector which gets you from one to the other, as this is in the direction of the line. If r is on the line, then r-a will be parallel to b.
7. No, no, I get you now, thanks. They always ask dumb questions on P6, I just prey for the straight forward 'find the inverse of this matrix' questions...
8. (Original post by Mat666)
No, no, I get you now, thanks. They always ask dumb questions on P6, I just prey for the straight forward 'find the inverse of this matrix' questions...
The good news is that the standardisation means you don't have to get as many real marks as usual to secure the same grade as the paper is harder (which is why I ended up with a better mark on P6 than D1 last year. )

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