In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook): http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up: ‹1, -1, 1› • ‹a, b, c› = 0 ‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0 b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c a = c - b = -2c
The computation, regardless of method, yields: ‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
You know that you have the Opposite side and the Adjacent side. therefore you would use Tangent (T= O/A) because you are finding the angle you would go Tan^-1(12÷20) =31° it is 12÷20 because O/A and the 12 is the opposite side therefore It would be O and 20 is the adjacent side therefore equaling 20 so O/A = 12/20 Your answer is 31°
