Proof of Sine Rule by vectors

Anyone know how to prove the Sine Rule using vectors? Please?

The vector diagram is labelled in standard notation for a triangle with sides a,b,c and opposite angles A,B,C.

Vector-wise,
c = a + b

Cross Products
axb = |a||b|sinC.n -> sinC = axb/(|a||b|.n)
bxc = |b||c|sinA.n -> sinA = bxc/(|b||c|.n)
cxa = |c||a|sinB.n -> sinB = cxa/(|c||a|.n)

bxc = bx(a+b) = bxa + bxb = -axb (bxb = 0)
cxa = (a+b)xa = axa + bxa = -axb (bxb = 0)

|a|/sinA = |a||b||c|.n / (bxc) = |a||b||c|.n / (-axb)
|b|/sinB = |b||c||a|.n / (cxa) = |a||b||c|.n / (-axb)

i.e. |a|/sinA = |b|/sinB etc.
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Reply 2

flyinghorse

Fermat

The vector diagram is labelled in standard notation for a triangle with sides.....

That's

. My faith in people and the internet has been redeemed. I've handed in my work before I read that, but now I'll look clever when I have my problem class tomorrow :smile:

I don't suppose you can also prove that the area is (1/2)|bxc|?

Reply 3

Fermat

flyinghorse

That's

. My faith in people and the internet has been redeemed. I've handed in my work before I read that, but now I'll look clever when I have my problem class tomorrow :smile:

I don't suppose you can also prove that the area is (1/2)|bxc|?

Show A = ½|bxc|sinC

Standard: A = ½|a||b|sinC

bxc = bx(a+b) = bxa + bxb = bxa
|bxc| = |bxa| = |axb|

from my earlier post,
axb = |a||b|sinC.n
|axb| = |a||b|sinC.|n|

but n is the unit vector and hence has a length of 1, i.e. |n| = 1

:. |axb| = |a||b|sinC

:. ½|bxc| = ½|axb| = ½|a||b|sinC

Edit: added in some extra lines to clarify the steps.