geometry evrybody?

Question

hey anybody know about triangle? that line AB is a projection form line Ac and BC... how's that suppose to work?!

Answer 1

You can do it using vectors and the "dot product," or else you can use the Law of Cosines. Let's do it the second way, using an example.

Let the three points be A(0,0), B(3,1), and C(5,6). This is a "scalene triangle," with angle ABC greater than 90 degrees. (I made it that way to be more general in what we're about to do.)

The three segment lengths are AB = sqrt(3^2 + 1^2) = sqrt 10; AC = sqrt(5^2 + 6^2) = sqrt 61; and BC = sqrt(2^2 + 5^2) = sqrt 29.

We're going to need angles CAB and CBA. These are the "base angles" of this triangle we'll need for the projection. Using the Law of Cosines:

BC^2 = AB^2 + AC^2 - 2 AB AC cos CAB
cos CAB = (AB^2 + AC^2 - BC^2) / (2 AB AC)
cos CAB = (10 + 61 - 29) / [2 sqrt(10*61)] = 21 / sqrt 610 = 0.850265
Angle CAB = 31.76 degrees

Using the same method:

cos CBA = (10 + 29 - 61) / [2 sqrt(10*29)] = -11 / sqrt 290 = -0.645942
Angle CBA = 130.24 degrees

Although we don't need it, we now know

angle ACB = 180 - 31.76 - 130.24 = 18.00 degrees

Now that we've done all the hard work, the projection itself is easy. The projection formula is:

AB = AC cos CAB + BC cos CBA (Equation 1)

AB = 0.850265 sqrt 61 + (-0.645942) sqrt 29
AB = 6.64078 - 3.47851 = 3.16228 (Answer)

which is the same as AB = sqrt 10 = 3.16228

Equation 1 above is the projection formula you're looking for. We did this problem the hard way, having to figure out the two base angles using the Law of Cosines. Usually the projection problem is a lot easier, since they give you the base angles right away, in which case you can use Equation 1 directly.

Answer 2

if they intersect they form an triangle