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# p6 vectors watch

1. find an equation in the form r.n =p which contains the line l and the point with position vector a where

l : r =-2i +3j -k + t (2i -j -3k ), a =-3i +j 2k

ne1 help me on dis plz?
2. r = (-2,3,-1) + t(2,3,-1)
the point (-2,3,-1) is on the line (at t=0), so a line from that point to a = (-3,1,2) is:

(-3,1,2)-(-2,3,1) = (-1,-2,1)

n is the vector perpendicular to both (-1,-2,1) and (2,3,-1)
=> n = (-1,-2,1) x (2,3,-1) = (-1,1,1)

p = (-2,3,-1) . (-1,1,1) = 4

=> r.(-i+j+k) = 4
3. no un4tun8ly this isnt the right answer

the answer is something like r.(9i+3j+5k) = -14 or something like that
the 9 and 14 are definitly right
4. I found the mistake (highlighted!). Just a minus sign. shows you how easy it is to make mistakes ( )

r = (-2,3,-1) + t(2,3,-1)
the point (-2,3,-1) is on the line (at t=0), so a line from that point to a = (-3,1,2) is:

(-3,1,2)-(-2,3,-1) = (-1,-2,3)

n is the vector perpendicular to both (-1,-2,3) and (2,3,-1)
=> n = (-1,-2,3) x (2,3,-1) = (-7,5,1)

p = (-2,3,-1) . (-7,5,1) = 28

=> r.(-7i+5j+k) = 28

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Updated: April 26, 2004
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