Well, let's use the Euclidean algorithm:
164 = 4×37 + 16
37 = 2×16 + 5
16 = 3×5 + 1
Then working backwards,
1 = 16 - 3×5
1 = 16 - 3×(37 - 2×16) = 7×16 - 3×37
1 = 7×(164 - 4×37) - 3×37 = 7×164 - 31×37
so that x = 7 and y = -31.
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:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
The greatest number of teams Leo can create is 2
<em><u>Solution:</u></em>
Given that,
There are 32 point guards and 50 posts In Leo's basketball league
From given,
Number of point guards = 32
Number of posts = 50
Leo must include all players on a team and wants each team to have the same number of point guards and the same number of posts
To find the greatest number of teams Leo can create, we should find the greatest common factor of 32 and 50
When we find all the factors of two or more numbers, and some factors are the same, then the largest of those common factors is the Greatest Common Factor
<em><u>Greatest Common Factor of 32 and 50:</u></em>
The factors of 32 are: 1, 2, 4, 8, 16, 32
The factors of 50 are: 1, 2, 5, 10, 25, 50
Then the greatest common factor is 2
Thus the greatest number of teams Leo can create is 2
