1. X, Y X axis Y axis Let’s just start with a point on a plane surface like this sheet of paper. Now coordinate “x” describes how far to the right, and “y” describes how far up the page the point is. So we’ll put a random point {X,Y} somewhere out in the x-y plane. Now we’ve located the point in a two-dimensional coordinate system. The origin of this coordinate system is where the two axes cross. Call that {0, 0} where x and y are both zero. 0, 0

2.  y x You could also draw a straight arrow from the origin of the coordinate system out to that point. The arrow representation of that point {X,Y} is a vector.

3.  y x (a) (b) (c) This is interesting because now the point we started with, which has become a vector, looks like a right triangle with side lengths (a) and (b), and hypotenuse (c). So the length of the hypotenuse is also the length of the vector and from trigonometry: Cos(  )= x/c and Sin(  )= y/c

4. Introduction to the “DOT product” The dot product is a beautiful and useful generalization of normal multiplication - only now you multiply vectors which are a bit more complicated. Since this is a new kind of multiplication, lets also give it a new symbol Lets call our vector from the last page just V for short. V really contains the two numbers x and y. The length of V squared is x 2 +y 2 or x*x+y*y from the Pythagorean formula. In words : “V squared, or V dot V, or VV, equals the length of the vector squared, and also equals the x component of V times the x component of V, plus the y component of V times the y component of V. Nothing new yet- remember this is a warm- up.

5.  y1y1 x1x1  y2y2 x2x2 Now expand the discussion to include two vectors V 1 and V 2 The dot product V 1V 2 equals the x component of V 1 times the x component of V 2 plus the y component of V 1 times the y component of V 2. In symbols: V 1V 2 =x 1 *x 2 +y 1 *y 2 Now this is becoming interesting.

6. Now lets assemble the pieces of that dot product ! Remember from trigonometry that the cosine of an angle equals the length of the adjacent side of a right triangle divided by the hypotenuse. The two hypotenuse lengths are just the lengths of the two vector V 1 and V 2. So…. x 1 *x 2 = cos(  )*cos(  )*length(V 1 )*length(V 2 ) The same argument works for sines, so y 1 *y 2 = sin(  )*sin(  )*length(V 1 )*length(V 2 ) Putting these piece together gives: V 1V 2 =length(V 1 )*length(V 2 )*(cos(  )*cos(  )+sin(  )*sin(  )) If you remember the easy derivations of the angle addition formulae from tidbit 1, this comes to: V 1V 2 =length(V 1 )*length(V 2 )*cos(  -  ). The angle difference (  -  is just the angle included between the two vectors. So this is one of several useful applications of the dot product – finding the angle between two vectors.

7. x y 0, 0, 0 z A 3D example – tetrahedral bonds. A molecule of methane (CH 4 ) is sketched out below. The four hydrogens are colored red and the central carbon in black. If the four bonds are symmetrical, the hydrogens should occupy opposite corners of a cube. Lets find the central angle between any pair of bonds

8. x y z Center the cube so that the carbon sits at the origin of the 3 dimensional coordinate system. Since the actual size of the cube won’t matter, lets just give it a side length of 2. Then the vector from the origin to the top left hydrogen ends at x = -1, y = 1, z = -1. So V 1 = {-1, 1, -1}. Doing the same for the hydrogen at the top right gives us V 2 ={1, 1, 1}. As before V 1V 2 =x 1 *x 2 +y 1 *y 2 +z 1 *z 2. Plugging the numbers in gives us V 1V 2 = -1. This also equals length(V 1 )*length(V 2 )*cos(  ). The result is cos(  )= -1/3 so  (the angle between bonds) is about 109.5 degrees. The vector lengths simplify since the two are equal. That leads to length 2 = V 1V 1 = 3