Dot product, is (a.b).(a.b)=(a.a).(b.b) ?

[lemd]

Hi,

As I remember, dot product is commutative, and so (a.b).(a.b) = (a.a).(b.b)
But when I apply to simple vectors it is all wrong, e.g:
a = (2, 2, 0)
b = (1, 0, 0)

(a.b).(a.b) = 2.2 = 4
(a.a).(b.b) = 8.1 = 8

Why are they different? Pls explain for me

Thanks

 

[AlephZero]

That is not what "commutative" means. It meas that (a.b) = (b.a)

It should be obvious that (a.b).(a.b) is not equal to (a.a).(b.b) in general. For example (a.b) can be 0 when (a.a) and (b.b) are both greater than 0.

 

[Mark44]

Also, your notation is misleading -- "(a.b).(a.b)". You are using a period (.) to indicate two different types of multiplication: the dot or scalar product that is defined for two vectors, and ordinary multiplication of real numbers.

Slightly better notation would be (a.b)(a.b), with nothing shown for the real number multiplication.

Even better would be to use a dot for the dot product, for which some simple LaTeX can be used: (a $\cdot$ b)(a $\cdot$ b).

 

[lemd]

Many thanks

 

[CellCoree]

for your question. The dot product is indeed commutative, which means the order of the vectors does not matter. However, in your example, you are not calculating the dot product correctly. The dot product is calculated by multiplying the corresponding components of the vectors and then adding them together. So for (a.b), it would be (2*1) + (2*0) + (0*0) = 2. Therefore, (a.b).(a.b) = 2*2 = 4, and (a.a).(b.b) = 8*1 = 8, which are the same. I hope this helps clarify things for you.