a.b=a bcosq
By Anonymous on Friday, April 20, 2001 - 04:09 pm :

Why is a.b=|a||b|cosq??
I really don't know what it means, can someone please explain it? and also prove it?

By Dan Goodman (Dfmg2) on Saturday, April 21, 2001 - 01:05 am :
Hi there. The reason for that formula is actually because we define a.b=|a||b|cos(q), and we derive the formula that you are familiar with from that. The angle q here is the angle between the vectors and |a| and |b| are the lengths of the vectors a and b.

If we assume that a.b=|a||b|cosq, can we work out a.(b+c)? Suppose that a=i (the unit vector in the x-direction), then |a|=1 and |b|cosq = bx, the x-coordinate of b (just draw a picture). Also, a.c=cx (x-coord of c) if c is another vector, and a.(b+c)=(b+c)x = bx+cx. So, we have proved that a.(b+c)=a.b+a.c if a=i the unit vector in the x-direction.

Suppose that a=R i, where R is a positive real number. Then we would have that a.b=|a||b|cosq = R|b|cosq = R(i.b). Obviously we also have a.c=R i.c and a.(b+c)=R i.(b+c), and so we have that a.(b+c)=a.b+a.c if a=R i.

Finally, we show that if you rotate both a and b some angle a about an axis v to get vectors a ' and b ' , then a.b=a ' .b ' . But this is obvious, because rotations do not change the lengths |a|=|a ' | and |b|=|b ' | or the angle (if q ' is the angle between a ' and b ' , then q ' =q since they have both been rotated by the same amount). Since we can always rotate avector a to a vector R i (just rotate it by 180 degrees about the vector R i+a), we have that a.(b+c)=a ' .(b+c) ' =a ' .(b ' +c ' ) = a ' .b ' +a ' .c ' =a.b+a.c (where a ' , b ' , (a+b) ' mean the vectors a, b, (a+b) rotated by the same amount so that a ' =R i).

OK, so we have that a.(b+c)=a.b+a.c for any vectors a, b and c. We also know that b.a=a.b, because the lengths and the angles are the same.

Now, what is (x i+y j+z k).(u i+v j+w k)? (Where i, j, k are the unit vectors in the x, y, z directions.) Well, if we let a=x i+y j+z k, b=u i and c=v j+w k, we get a.(b+c)=a.b+a.c, and so (x i+y j+z k).(u i+v j+w k)=(x i+y j+z k).(u i)+(x i+y j+z k).(vj+wk). Expanding the second term out we get (x i+y j+z k).(u i)+(x i+y j+z k).(v j)+(x i+y j+z k).(w k)=¼ = x u+y v+z w, which is the formula you know for the dot product.

Summary: We defined a.b=|a||b|cosq, we proved (from this) that a.(b+c)=a.b+a.c and then proved (from this and the obvious facts that i.j=0, i.i=1 and so on) a.b=ax bx +ay by +az bz.

Did that make sense?

By Woon Khang Tang (P3742) on Saturday, April 21, 2001 - 03:41 am :

Thank you!!! Even though I don't really understand at first glance, I'll print it out and read it again until I understand. I'm sure I'll understand, and a million thanks for your detail explanation. I'm really desperate after I've found through dozens of books and my teacher didn't explain why. I'm really surprise when I asked my friend and they told me just memorize the formula as long as you know how to apply the formula, it's ok. I really hate to memorize formula without understanding it and proving it. Without understanding the formula, when I apply the formula, it's like you can find the right answer easily, but you don't know what the heck are you doing, and that's really really stupid!!!

By Michael Doré (Michael) on Wednesday, April 25, 2001 - 02:27 pm :

Your textbook might alternatively define a .b as a1 b1 + a2 b2 + a3 b3 where (a1 ,a2 ,a3 ) and (b1 ,b2 ,b3 ) are the Cartesian components of a and b . Note that a and b might be only 2-D - in this case we can set a3 = b3 = 0 and it all works the same way, because we confine ourselves to a 2-D plane (the XY plane in fact).

Anyway, we form a triangle, OAB where O is the origin and the vector OA is the vector a and the vector OB is the vector b .

If we let | | denote length then by Pythagoras, we know:

|OA|2 = a1 2 + a2 2 + a3 2

(because a1 ,a2 ,a3 are the Cartesian components of the A, because A has position vector a )

Likewise:

|OB|2 = b1 2 + b2 2 + b3 2

Also we know that the vector AB has components (b1 -a1 ,b2 -a2 ,b3 -a3 ), so we get:

|AB|2 = (b1 - a1 )2 + (b2 - a2 )2 + (b3 - a3 )2

Expand this:

|AB|2 = b1 2 + b2 2 + b3 2 + a1 2 + a2 2 + a3 2 - 2a1 b1 - 2a2 b2 - 2a3 b3

But using our results above, and the definition of a .b :

|AB|2 = |OA|2 + |OB|2 - 2a .b

But by the cosine rule:

|AB|2 = |OA|2 + |OB|2 - 2|OA||OB|cos(AOB)

Comparing these two equations we obtain:

a .b = |OA||OB|cos(AOB)

By Kerwin Hui (Kwkh2) on Wednesday, April 25, 2001 - 10:16 pm :
The formula a.b=|a||b| cosq is a definition for the angle between 2 vectors when you are in higher dimensions.

Kerwin