# Vector help?Watch

#1
I'm stuck on this vector question and my method doesnt work :/

(iii) If the point S has position vector OS (3i -8j -7k) use a vector method to show that the points P, Q and S all lie on the same straight line.
OP (1i 2j 3k)
OQ (2i -3j -2k)
I thought if it lies on the same straight line then there would be a scalar k that would make it so multiplying any of the vectors by k I would get the other ones :/
0
4 years ago
#2
(Original post by saberahmed786)
I'm stuck on this vector question and my method doesnt work :/

I thought if it lies on the same straight line then there would be a scalar k that would make it so multiplying any of the vectors by k I would get the other ones :/
You need to find the vectors and check using your approach. These are the vectors that lie along the line (if that's the case). You have been given position vectors of the points i.e. the vectors that translate the origin to those points.

This will make more sense if you draw a diagram - you should always try to draw a diagram when doing vector problems (or any other problem, in fact).
0
#3
(Original post by atsruser)
You need to find the vectors and check using your approach. These are the vectors that lie along the line (if that's the case). You have been given position vectors of the points i.e. the vectors that translate the origin to those points.

This will make more sense if you draw a diagram - you should always try to draw a diagram when doing vector problems (or any other problem, in fact).
I get
pq = (i -5j -5k)
qs = (i -11j -9k)
sp = (-2i 10j 10k)

I can see that pq and sp are the same because it's multiplied by -2 but what about qs?
0
4 years ago
#4
(Original post by saberahmed786)
I get
pq = (i -5j -5k)
qs = (i -11j -9k)
sp = (-2i 10j 10k)

I can see that pq and sp are the same because it's multiplied by -2 but what about qs?
You need to check your working for qs - I don't think that's right at all
0
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