Proof involving dot product of 3 vectors

Question

I would like help in finding an a resolution to a problem I have in a proof involving vectors.

I will present my solution as well and explain where I believe the problems are after the formulation of the problem.

Given a definition of a dot product as:

$x \cdot y$ = $ x_1 y_y+\cdots + x_ny_n$

Now assume we have mutually orthogonal vectors $f_1,\cdots, f_n$

and the representation: $w= a_1f_1+\cdots+a_mf_m$

where $a_1,\cdots,a_m$ are scalars.

Prove that:

$\frac{w\cdot f_i}{f_i \cdot f_i}=a_i$ for every $i=1,\cdots,m$

My solution: Let

$f_i=(f_1,\cdots, f_m)$ and

$a_i=(a_1,\cdots, a_m)$

So now I have a vector of scalars $a_i$ and a vector of vectors $f_i$ so I can do the following:

1. $w\cdot f_i=(a_i \cdot f_i)\cdot f_i $

2. $w\cdot f_i=f_i(a_1f_1+\cdots +a_mf_m)$

3. $w\cdot f_i=(f_1(a_1f_1+\cdots +a_mf_m),\cdots,f_m(a_1f_1+\cdots +a_mf_m))$
4. $w\cdot f_i=(f_1(a_i\cdot f_i),\cdots,f_m(a_i\cdot f_i))$
5. $w\cdot f_i=f_i(a_i\cdot f_i)$
6. $w\cdot f_i=a_i(f_i \cdot f_i)$
7. $\frac{w\cdot f_i}{f_i \cdot f_i}=a_i$

So my first potential issue is the formulation of $a_i$ and $f_i$. I've formulated them as I thought they should be but it's possible I've misunderstood and they're not really a vector of scalars and vector of vectors respectively. In which case my proof immediately falls apart.

The second issue is steps 5 and 6. I know that there is no associative law between dot products but I thought that because $f_i$ consists of orthogonal vectors, it would be equivalent because the orthogonal vectors would cancel out leaving on the cases where $f_i \cdot f_i$.

Lastly is it possible that $w$ in this case is actually referring to a specific term in the summation, like $w_k$ for example. If the proof is to demonstrate that a $w$ at some index is some a $a_i$ multplied by some vector $f_i$ in which case the proof would make sense since $a_i$ would be a scalar instead of a vector of scalars as I originally thought.

Thank you for any assistance in advance

Answer 1

I don't understand why you are complicating things when it can be done so simply.

$$w\cdot f_i=(a_1f_1+a_2f_2+\cdot \cdot \cdot +a_mf_m)\cdot f_i$$

$$w\cdot f_i=a_1(f_1\cdot f_i)+a_2(f_2\cdot f_i)+\cdot \cdot \cdot +a_m(f_m \cdot f_i)$$

Now because $f_1, \cdot \cdot \cdot ,f_m $ are mutually orthogonal, $f_i \cdot f_m =0$ for $i \ne m$

Also note that dot product of a vector with itself that is $f_i \cdot f_i=||f_i||^2$, so we have:

$$\frac{w\cdot f_i}{f_i \cdot f_i}=\frac{a_1(f_1\cdot f_i)+a_2(f_2\cdot f_i)+\cdot \cdot \cdot +a_m(f_m \cdot f_i)}{f_i \cdot f_i }$$

$$\implies \frac{w\cdot f_i}{f_i \cdot f_i}=\frac{0+0+\cdot \cdot a_i(f_i \cdot f_i)+\cdot \cdot +0+0}{f_i \cdot f_i}=a_i\frac{||f_i||^2}{||f_i||^2}=a_i$$

Answer 2

The dot product is linear in its arguments, meaning that the dot product of a linear combination of vectors (with another vector) is the linear combination of the respective dot products.

Hence,

$$w\cdot f_k=a_1(f_1\cdot f_k)+a_2(f_2\cdot f_k)+\cdots a_1(f_k\cdot f_k)+a_n(f_k\cdot f_k) \\=a_1\,0+a_2\,0+\cdots a_k(f_k\cdot f_k)+\cdots a_n\,0.$$