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Probably very simple vectors question

Hi, I've got what should be, on the surface, a very simple vectors question but no matter what I do, I can't seem to make any headway! If anyone could give me a hint to start me off, that would be fantastic.

If a, b and c are coplanar vectors related by la + mb + n c=0, where l, m, n are all non-zero, show that the conditions for the points da, eb and fc to be colinear is:
ld+me+nf=0 \frac{l}{d} + \frac{m}{e} + \frac{n}{f} = 0

Thanks alot,

Masiyi

Reply 1

Oof. Now, my initial thought (if you'll excuse the lazy notation; I can't be arsed) is the following. We know that da-eb = k(fc-eb) for some scalar k. Rearranging, we get da + e(k-1)b + (-fk)c = 0, which looks to me surprisingly similar to the given equation in l, m, n. We can deduce that l/d = e(k-1)/m = -fk/n, and work out k from that... and that's as far as I've got.

Looks like it should come out from there but it certainly doesn't look easy. There's almost certainly a better way.

Reply 2

generalebriety
Oof. Now, my initial thought (if you'll excuse the lazy notation; I can't be arsed) is the following. We know that da-eb = k(fc-eb) for some scalar k. Rearranging, we get da + e(k-1)b + (-fk)c = 0, which looks to me surprisingly similar to the given equation in l, m, n. We can deduce that l/d = e(k-1)/m = -fk/n, and work out k from that... and that's as far as I've got.

Looks like it should come out from there but it certainly doesn't look easy. There's almost certainly a better way.
One way (and yes, it's a full solution... :rolleyes: )

Spoiler



Note to LaTeXer's - as I discovered, you'll encounter trouble if you end up with the product def when typing this up. Try typing d{}ef to stop it being treated as a special reserved word.

Edit: having had a minute to think after the LaTeX hell, this feels like total overkill for this problem - there must be something simpler.

Reply 3

You're an absolute genius! And it seems that it was slightly more challenging that I expected!

Once again, thanks alot!