Show that CP is independent of the values of r and q and evaluate it (vectors)

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TSR360
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Question (part c): http://prntscr.com/s68muw


vector AC = 8i -4, vector BP = (r - 6)i + (q + 3)j and vector CP = (r - 8)i + (q + 4)j


How do I show CP is independent of p and r? If that was possible then r and q wouldn't be in the CP vector unless they were both equal to 0 (which they're not). And how do I 'evaluate' a vector?? I'm so confused...
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Pangol
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(Original post by TSR360)
Question (part c): http://prntscr.com/s68muw


vector AC = 8i -4, vector BP = (r - 6)i + (q + 3)j and vector CP = (r - 8)i + (q + 4)j


How do I show CP is independent of p and r? If that was possible then r and q wouldn't be in the CP vector unless they were both equal to 0 (which they're not). And how do I 'evaluate' a vector?? I'm so confused...
You haven't yet used the information that AP = 2BP.

Note that AP and BP do not have arrows over them, just like CP in (c). This means we are talking about the magnitude of the vectors, not the vectors themselves.
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(Original post by Pangol)
You haven't yet used the information that AP = 2BP.

Note that AP and BP do not have arrows over them, just like CP in (c). This means we are talking about the magnitude of the vectors, not the vectors themselves.
I still don’t see how that helps...
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Pangol
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(Original post by TSR360)
I still don’t see how that helps...
Start by writing an equation, in terms of r and q, for what it means to have AP = 2BP. It will have lots of terms and you won't be able to simplify it much, but do it anyway and then put it to one side.

Then work out an expression for CP, again in terms of r and q. The equation you worked on in the first step should suggest itself to you at this point.

It will be easiest if you do all this not in terms of the actual magnitudes of the vectors, but in terms of the squares of the magnitudes of the vectors (just to save you having to keep wrapping everything in huge square roots).
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(Original post by Pangol)
Start by writing an equation, in terms of r and q, for what it means to have AP = 2BP. It will have lots of terms and you won't be able to simplify it much, but do it anyway and then put it to one side.

Then work out an expression for CP, again in terms of r and q. The equation you worked on in the first step should suggest itself to you at this point.

It will be easiest if you do all this not in terms of the actual magnitudes of the vectors, but in terms of the squares of the magnitudes of the vectors (just to save you having to keep wrapping everything in huge square roots).
Like this? http://prntscr.com/s6apap
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Pangol
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(Original post by TSR360)
Like this? http://prntscr.com/s6apap
I don't really follow that. What are the lower case vectors? Where have you used the fact that AP = 2BP? In fact, where have you worked out any magnitudes at all?
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(Original post by Pangol)
I don't really follow that. What are the lower case vectors? Where have you used the fact that AP = 2BP? In fact, where have you worked out any magnitudes at all?
I’ve re-written the displacement vectors so I can re-arrange things better. I turned AP = 2BP into |p - a| = 2|p - b| then substituted p - a in the CP vector with 2(p - b) so I can solve the vector simultaneously to find r and q and then re-write the final vector in terms of A, B and C
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(Original post by TSR360)
I’ve re-written the displacement vectors so I can re-arrange things better. I turned AP = 2BP into |p - a| = 2|p - b| then substituted p - a in the CP vector with 2(p - b) so I can solve the vector simultaneously to find r and q and then re-write the final vector in terms of A, B and C
You cannot turn AP = 2BP into a vector equation. It is telling you that the magnitude of the vector AP is twice the magnitude of the vector BP. It is not telling you that the vector AP is twice the vector BP.
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