Undergrad Parallel and perpendicular components

[Portuga]

TL;DR

Consider the following vectors of $\mathbb{V}^{3}$: $u=\left(2,2,2\right)$
and $v=\left(3,3,1\right)$. If $w=\left(-5,1,-1\right)$, decompose
the vector $v$ into a component in $W=\left[u,w\right]$ and a component
in $W^{\perp}$.

I first tried to use a method based on Gram Schmidt orthogonalization
method:
$$
v_{\parallel}=\left(v\ldotp\frac{u}{\left\Vert u\right\Vert }\right)\frac{u}{\left\Vert u\right\Vert }+\left(v\ldotp\frac{w}{\left\Vert w\right\Vert }\right)\frac{w}{\left\Vert w\right\Vert },
$$
and
$$
v_{\perp}=v-v_{\parallel}.
$$
Results were
$$
v_{\parallel}=\left(\frac{128}{27},\frac{50}{27},\frac{76}{27}\right)
$$
and
$$
v_{\perp}=\left(-\frac{47}{27},\frac{31}{27},-\frac{49}{27}\right).
$$
At first sight, things looked ok, because $v_{\parallel}+v_{\perp}=v$,
but when I performed the scalar product between $v_{\perp}$ and $u$
and $w$, both resulted non null:
$$
v_{\perp}\ldotp u=-\frac{130}{27}
$$
and
$$
v_{\perp}\ldotp w=\frac{35}{3}.
$$
This was totally unespected to me because $v_{\perp}$ was designed
to be perpendicular to the subspace spanned by $u$ and $w$, and
as so, these scalar products should result 0.

Then I tried a slight deviation from the previous method:
$$
v_{\parallel}=\left[v\ldotp\frac{\left(u+w\right)}{\left\Vert u+w\right\Vert }\right]\frac{\left(u+w\right)}{\left\Vert u+w\right\Vert }=\left(-\frac{3}{19},\frac{3}{19},\frac{1}{19}\right)
$$
and
$$
v_{\perp}=v-v_{\parallel}=\left(\frac{60}{19},\frac{54}{19},\frac{18}{19}\right).
$$
Again,
$$
v_{\perp}\ldotp u=\frac{264}{19}
$$
and
$$
v_{\perp}\ldotp w=-\frac{264}{19}.
$$
But, there was an advance:
$$
v_{\perp}\ldotp\left(u+w\right)=0.
$$
So, what am I doing wrong on all this? Am I missing something?

 

[Office_Shredder]

Gram Schmidt goes step by step constructing an orthogonal set of vectors, but it assumes you did the right thing on the previous vectors on each step. Here, u and w are not orthogonal, so you can't orthogonalize the third vector immediately. You could run the algorithm from scratch on the set of u,w,v and do two iterations.

 

[Portuga]

I should subtract first component, normalize and then subtract the other one, and normalize it?

 

[Office_Shredder]

I should subtract first component, normalize and then subtract the other one, and normalize it?

This is a little too vague for me to know if you're proposing the right thing.

 

[Steve4Physics]

@Portuga, can I add a few thoughts to what @Office_Shredder has said…

Thinking geometrically, $v_{\parallel}$ is simply the projection of $v$ onto the plane (P) spanned by $u$ and $w$. Not too hard to visualise.

Your equation
$v_{\parallel}=\left(v\ldotp\frac{u}{\left\Vert u\right\Vert }\right)\frac{u}{\left\Vert u\right\Vert }+\left(v\ldotp\frac{w}{\left\Vert w\right\Vert }\right)\frac{w}{\left\Vert w\right\Vert }$
would give you $v_{\parallel}$ if $u$ and $w$ were orthogonal - but they aren’t!

You can find an orthogonal basis for $[u,w]$ using the Gram Schmidt method. And then you can use the above equation but with the orthogonal basis vectors.

An alternative approach would be to start by finding the cross product $u \times w$. This is a vector perpendicular to P so it allows you to find $v_{\perp}$.

 

[Portuga]

ok, thank you all very much!

 

[SammyS]

ok, thank you all very much!

You're welcome.
What did you get for results, and what method did you use?

After giving help, we're often curious as to the outcome.

 

[Portuga]

I followed the trick of Gram Schmidt ortogonalization method to build an ortogonal basis $\left\{ u^{\prime},w^{\prime}\right\}$ for $W$ and the strategy of decomposing $v$ as $v_{\parallel}=v\ldotp u^{\prime}u^{\prime}+v\ldotp w^{\prime}w^{\prime}$ and $v_{\perp}=v-v_{\parallel}$. It worked like a charm! Thank you all very much.