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Fp3 june 2010 - vectors

HI ALL.
Can anyone please explain how to do part c of question 7 on vectors. I am a little lost on how they got to the answer in the mark scheme.
Hope revision is going well.

Thanks
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B_9710
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Original post by DORI
HI ALL.
Can anyone please explain how to do part c of question 7 on vectors. I am a little lost on how they got to the answer in the mark scheme.
Hope revision is going well.

Thanks


Which paper? Or post a link.
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DORI
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Original post by B_9710
Which paper? Or post a link.


https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202010%20QP%20-%20FP3%20Edexcel.pdf page 11

thanks
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B_9710
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For FP3 there is a standard formula that gives you the shortest distance from a point to a line. Alternatively you could find an equation of the line from N that is perpendicular to PR. Then using the fact that the line you found the equation when dot multiplied with the direction of PR will be 0 you can find the value of the parameter which enables you to easily find the distance.
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Original post by B_9710
Alternatively you could find an equation of the line from N that is perpendicular to PR. Then using the fact that the line you found the equation when dot multiplied with the direction of PR will be 0 you can find the value of the parameter which enables you to easily find the distance.


that's were I seem to be a little lost. I am trying to find the equation of the line from N to PR, but with no success! :frown:
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B_9710
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Original post by DORI
that's were I seem to be a little lost. I am trying to find the equation of the line from N to PR, but with no success! :frown:


Well PR=−5i−13j−3k PR= -5\mathbf{i}-13\mathbf{j}-3\mathbf{k} . So the parametric equation of the line PR, is r=6i+13j+5k+λ(5i+13j+3k) \mathbf{r} =6\mathbf{i}+13\mathbf{j}+5 \mathbf{k}+\lambda (5\mathbf{i}+13\mathbf{j}+3 \mathbf{k}) . (I'm using i, j k notation as it is easier to do on here).
So if we let point B (arbitrarily chosen) lie on PR, such that NC is perpendicular to PR, then we realise that the point C is the point on PR that is closest to N. If N lies on PR then it must have coordinates that satisfies the equation of PR. Using properties of the scalar product you should be able to find the value of λ \lambda and then find the distance.
Please just say if you want me to clarify any of this.
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Original post by B_9710
Well PR=−5i−13j−3k....λ(5i+13j+3k) PR= -5\mathbf{i}-13\mathbf{j}-3\mathbf{k} .... \lambda (5\mathbf{i}+13\mathbf{j}+3 \mathbf{k}) . (I'm using i, j k notation as it is easier to do on here).


should that not be lambda ( -5i -13j -3k).
Btw thanks so much for your help.
(edited 7 years ago)
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Original post by DORI
should that not be lambda ( -5i -13j -3k).
Btw thanks so much for your help.


Because λ \lambda is just a constant, it could be positive or negative, so it doesn't matter what the direction vector is as long as it is scaled correctly. It could be (-10i -26j 6k ) if you wanted it to be, you would just get different λ \lambda values so it would't matter at all.
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Original post by B_9710
Because λ \lambda is just a constant, it could be positive or negative, so it doesn't matter what the direction vector is as long as it is scaled correctly. It could be (-10i -26j 6k ) if you wanted it to be, you would just get different λ \lambda values so it would't matter at all.


ok I see.

I've finally got it. :biggrin:

Thanks

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