Vector proof

Not too sure if I show post this question in A level or undergrad, hopefully someone can answer it for me.
Proof that

\mathbf{a}

.

\mathbf{b}

=

\frac{1}{4}

(|\mathbf{a}+\mathbf{b}|^2)

-

(|\mathbf{a}-\mathbf{b}|^2)

My attempt was trying to generalise the vectors a and b slightly such that

\mathbf{a}

= (a1, a2, a3) and

\mathbf{b}

= (b1, b2, b3) and start on the right hand side and eventually it will become a1b1+a2b2+a3b3 which is the same as a dot b but I was wondering does dot product apply more than 3 dimensions, for instance, should I make

\mathbf{a}

= (A1,A2,A3,A4...An) instead?

(edited 5 years ago)

Original post by Iconic_panda

Not too sure if I show post this question in A level or undergrad, hopefully someone can answer it for me.
Proof that

\mathbf{a}

.

\mathbf{b}

=

\frac{1}{4}

(|\mathbf{a}+\mathbf{b}|^2)

-

(|\mathbf{a}-\mathbf{b}|^2)

My attempt was trying to generalise the vectors a and b slightly such that

\mathbf{a}

= (a1, a2, a3) and

\mathbf{b}

= (b1, b2, b3) and start on the right hand side and eventually it will become a1b1+a2b2+a3b3 which is the same as a dot b but I was wondering does dot product apply more than 3 dimensions, for instance, should I make

\mathbf{a}

= (A1,A2,A3,A4...An) instead?

If you are going to take such an approach, then yes you need to generalise to arbitrary dimension

n

by saying that

\mathbf{a} = (a_1, a_2 , \ldots , a_n)

I assume you mean

\mathbf{a} \cdot \mathbf{b} = \frac{1}{4} \left( |\mathbf{a} + \mathbf{b}|^2 - |\mathbf{a} - \mathbf{b}|^2 \right)

.

You can say that

|\mathbf{a} + \mathbf{b}|^2 = (a_1 + b_1)^2 + (a_2 + b_2)^2 + \ldots + (a_n + b_n)^2

|\mathbf{a} - \mathbf{b}|^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + \ldots + (a_n - b_n)^2

Then clearly

Unparseable latex formula:

\begin{aligned} |\mathbf{a} + \mathbf{b}|^2 - |\mathbf{a} - \mathbf{b}|^2 & = [(a_1 + b_1)^2 - (a_1 - b_1)^2] + [(a_2 + b_2)^2 - (a_2 - b_2)^2] + \ldots + [(a_n + b_n)^2 - (a_n - b_n)^2] \\ & = ???

Finish it off from there.

(edited 5 years ago)

Reply 2

DFranklin

18

Original post by Iconic_panda

Not too sure if I show post this question in A level or undergrad, hopefully someone can answer it for me.
Proof that

\mathbf{a}

.

\mathbf{b}

=

\frac{1}{4}

(|\mathbf{a}+\mathbf{b}|^2)

-

(|\mathbf{a}-\mathbf{b}|^2)

My attempt was trying to generalise the vectors a and b slightly such that

\mathbf{a}

= (a1, a2, a3) and

\mathbf{b}

= (b1, b2, b3) and start on the right hand side and eventually it will become a1b1+a2b2+a3b3 which is the same as a dot b but I was wondering does dot product apply more than 3 dimensions, for instance, should I make

\mathbf{a}

= (A1,A2,A3,A4...An) instead?

In full generality, the dot product applies to things where "dimensions" (in the way you are thinking about them) doesn't even make sense. (For example, for real continuous functions on [0,1] you could define

f.g = \int_0^1 f(x)g(x)\,dx

).

If you want to do this for general vectors, I would use the fact that

|\mathbf{u}|^2 = \mathbf{u}\cdot \mathbf{u}

and the distributive law for the dot product. It's a very short proof this way.

Edit: I see you said this was possibly a "undergrad" question: certainly for an undergrad proof, I would do what I suggest instead of writing a = (a1, a2, ..., an) etc.

(edited 5 years ago)

Vector proof

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