Linear Algebra - Projecting a vector onto another

[TheTraveler]

Hi all!
I was studying a demonstration about vectors projections and it starts with this image:



The demonstration starts solving by v_|| and it says that:

$$\mathbf{v}_\| = \mathbf{n}(\| \mathbf{v}_\|\| / \| \mathbf{n}\|)$$
and this is the unclear step.
I tried to unroll this step (which is not explained in the demonstration) and I thought about using the dot product:

$$\mathbf{n} \cdot \mathbf{v}_\| = \|\mathbf{n}\| * \|\mathbf{v}_\|\| * \cos\theta$$
and, because the two vectors are parallels, the angle is 0 and the cosine is 1, so it results:

$$\mathbf{n} \cdot \mathbf{v}_\| = \|\mathbf{n}\| * \|\mathbf{v}_\|\|$$
but from here on out I don't really understand how to proceed to obtain the first step. Any suggestion? I know it's probably a stupid thing but I cannot figure it out ?

 

[Deleted member 4993]

Hi all!
I was studying a demonstration about vectors projections and it starts with this image:

View attachment 32627

The demonstration starts solving by v_|| and it says that:

$$\mathbf{v}_\| = \mathbf{n}(\| \mathbf{v}_\|\| / \| \mathbf{n}\|)$$
and this is the unclear step.
I tried to unroll this step (which is not explained in the demonstration) and I thought about using the dot product:

$$\mathbf{n} \cdot \mathbf{v}_\| = \|\mathbf{n}\| * \|\mathbf{v}_\|\| * \cos\theta$$
and, because the two vectors are parallels, the angle is 0 and the cosine is 1, so it results:

$$\mathbf{n} \cdot \mathbf{v}_\| = \|\mathbf{n}\| * \|\mathbf{v}_\|\|$$
but from here on out I don't really understand how to proceed to obtain the first step. Any suggestion? I know it's probably a stupid thing but I cannot figure it out ?

In my opinion, you have the correct picture in your mind. Dot product IS the projection of one vector onto the other.

Projection of A onto B = $\displaystyle \frac{A \cdot B}{||B||}$

 

[TheTraveler]

In my opinion, you have the correct picture in your mind. Dot product IS the projection of one vector onto the other.

Projection of A onto B = $\displaystyle \frac{A \cdot B}{||B||}$

Thank you for the reply!
So in my case:

$$\mathbf{proj_nv_\|} = \mathbf{v_\|} = \frac {\mathbf{v}_\| \cdot \mathbf{n}}{{\|\mathbf{n}\|^2}} \mathbf{n} = \frac{\|\mathbf{v}_\|\| \|\mathbf{n}\|}{\|\mathbf{n}\|^2}\mathbf{n} = \frac{\|\mathbf{v}_\|\|}{\|\mathbf{n}\|}\mathbf{n}$$
If so, the thing that got me confused is that the demonstration starts using the formula it wants to demonstrate but without making it explicit

 

[blamocur]

I believe you are missing this formula: $$\mathbf v_{||} = \mathbf n \frac{\mathbf v\cdot \mathbf n}{||\mathbf n||^2}$$

 

[Dr.Peterson]

If so, the thing that got me confused is that the demonstration starts using the formula it wants to demonstrate but without making it explicit

It may be helpful if you show us everything it says up to the point you are asking about, so we can be sure we understand the context, and possibly correct some misunderstanding about the flow of thought. We need to know what is known at this point in the argument.

 

[TheTraveler]

Ok, these are the concepts explained before the projection paragraph:
- Zero vector
- Negating a vector
- Magnitude
- Vector multiplication by a scalar
- Normalization
- Vector addition and subtraction
- Distance formula
- Dot product

After the last point, the book starts to explaine the projection:




Only at the end it shows the projecting formula, so I miss from where the demonstration starts

 

[Steven G]

V_|| and n are in the same direction.
So V_|| is a scalar multiple of n. That is V_|| = kn, for some scalar n.

Now vector n has some length and if we divide n by its length, then we get a vector of length 1 in the direction of vector n (and vector V_||). We can accomplish this as n/|n|. So we have a vector n/|n| whose length is 1. But we do not want the length in the end to be 1, we want it to be the length of V_||, namely |V_|||. Now (n/|n|) |V_||| = (|n|/ |V_|||)n has a length of |V_|||. So V = (|n|/ |V_|||)n

 

[TheTraveler]

V_|| and n are in the same direction.
So V_|| is a scalar multiple of n. That is V_|| = kn, for some scalar n.

Now vector n has some length and if we divide n by its length, then we get a vector of length 1 in the direction of vector n (and vector V_||). We can accomplish this as n/|n|. So we have a vector n/|n| whose length is 1. But we do not want the length in the end to be 1, we want it to be the length of V_||, namely |V_|||. Now (n/|n|) |V_||| = (|n|/ |V_|||)n has a length of |V_|||. So V = (|n|/ |V_|||)n

Hi Steven, thank you for the reply!
I knew I was missing a simple thing: so it's just a combination of normalization and multiplication by a scalar.
I was so focused on the dot product that I didn't think about anything else ?
Thank you all!

 

[pka]

I was studying a demonstration about vectors projections and it starts with this image:
View attachment 32627

If each of ${\bf{v}}~\&~{\bf{n}}$$$ is a non-zero vector then ${\bf{v}}$ can be resolved into the
sum of two vectors, one, ${\bf{v_{||}}}$, parallel to ${\bf{n}}$ the other,${\bf{v_{\bot}}}$, perpendicular to ${\bf{n}}$.
This is taken from Introduction to Vector Analysis by Davis & Snider
that is to say that ${\bf{v}}={\bf{v_{||}}}+{\bf{v_{\bot}}}$ moreover $\bf{v}=\alpha{\bf{v_{||}}}~\text{ and }~\bf{v}{\bf{\cdot v_{\bot}}}=0$.
If the notation $\|\bf{v}\|$ stands for the length of $\bf{v}$$$$$$$ then $\|\bf{v}_{\|}\|=\dfrac{|\bf{v}\cdot\bf{n}|}{\|\bf{n}\|}$ and $\|\bf{v}_{\bot}\|=\dfrac{\|\bf{v}\times\bf{n}\|}{\|\bf{n}\|}$
$$$$$$$$

 

[Steven G]

Hi Steven, thank you for the reply!
I knew I was missing a simple thing: so it's just a combination of normalization and multiplication by a scalar.
I was so focused on the dot product that I didn't think about anything else ?
Thank you all!

Just understand that what you call normalization is also a scalar. That scalar is (1/|n|)

 

[Steven G]

@Subhotosh Khan in post number 7 above, on the 2nd line can you please change that very last letter, n, to a k. Also in my last line, please replace (|n|/ |V||) which is written twice with (|V||/|n|)
Then remove this post.
Thank you!