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merged 1 commit into from
Aug 9, 2025

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syed0369
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@syed0369 syed0369 commented Jul 9, 2025

To see the next important property we should take a look at the set of points $\mathbf r$ for which $\mathbf r\cdot \mathbf a = C$ for some fixed constant $C$.
You can see that this set of points is exactly the set of points for which the projection onto $\mathbf a$ is the point $C \cdot \dfrac{\mathbf a}{|\mathbf a|}$ and they form a hyperplane orthogonal to $\mathbf a$.

I guess for $\mathbf r\cdot \mathbf a = C$ the point of projection is $\mathbf r \cos \theta$ which is of magnitude $C/|\mathbf a|$ and has direction of $\hat{a}$ (unit vec along ${\mathbf a}$) so the point is $C \cdot \dfrac{ \hat{\mathbf{a}}}{|\mathbf a|}$ which evaluates to the point $C \cdot \dfrac{\mathbf a}{|\mathbf a|^2}$

@mhayter
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mhayter commented Aug 8, 2025

Chatgpt disagrees: Great question — you're close, but the change you've made is not correct from a geometric and mathematical standpoint.

Let’s clarify:

✅ Original (correct) version:
You can see that this set of points is exactly the set of points for which the projection onto
𝑎
a is the point
𝐶

𝑎

𝑎

C⋅
∣a∣
a

,
and they form a hyperplane orthogonal to
𝑎
a.

This is correct, because:

The projection length is constant (equal to
𝐶
C), so the actual point on the line of
𝑎
a is at a distance of
𝐶
C units along the direction of
𝑎
a.

The unit vector in the direction of
𝑎
a is
𝑎

𝑎

∣a∣
a

, so scaling it by
𝐶
C gives you the fixed projection point.

❌ Your modified version:
... the projection onto
𝑎
a is the point
𝐶

𝑎

𝑎

2
C⋅
∣a∣
2

a

This is incorrect, because:

𝑎

𝑎

2
∣a∣
2

a

is not a unit vector — it's a scaled-down version of
𝑎
a.

This form appears in a different context: when you’re projecting a vector onto another vector, the formula for projection is:

proj
𝑎
(
𝑟
)

𝑟

𝑎

𝑎

2

𝑎
proj
a

(r)=
∣a∣
2

r⋅a

⋅a
But here you're describing the set of vectors whose dot product with
𝑎
a is constant, not projecting a specific vector onto another.

✅ TL;DR:
Keep the original form:

𝐶

𝑎

𝑎

C⋅
∣a∣
a

It is geometrically meaningful and correct — describing a point lying at a fixed distance
𝐶
C along the direction of
𝑎
a, which is how the hyperplane is constructed.

Let me know if you'd like a visual explanation of this with an example or diagram.

You said:
Yes

ChatGPT said:
Great! Let’s walk through a visual explanation of the statement:

“The set of points
𝑟
r such that
𝑟

𝑎

𝐶
r⋅a=C form a hyperplane orthogonal to
𝑎
a, and the projection of any such
𝑟
r onto
𝑎
a is the point
𝐶

𝑎

𝑎

C⋅
∣a∣
a

.”

🧠 Step-by-step Visual Intuition:

  1. What does
    𝑟

    𝑎
    =
    𝐶
    r⋅a=C mean?
    The dot product
    𝑟

    𝑎
    r⋅a is:


𝑟



𝑎


cos

𝜃
∣r∣⋅∣a∣⋅cosθ
where
𝜃
θ is the angle between
𝑟
r and
𝑎
a.

So fixing
𝑟

𝑎

𝐶
r⋅a=C means all such vectors
𝑟
r have the same projection onto
𝑎
a.

  1. Geometric meaning of constant projection
    Imagine you have a vector
    𝑎
    a, and you're projecting various points in space (vectors
    𝑟
    r) onto the direction of
    𝑎
    a. If all those projections land at the same spot on the line of
    𝑎
    a, those points form a flat surface (or line, in 2D) that is perpendicular to
    𝑎
    a.

That surface is called a hyperplane orthogonal to
𝑎
a.

  1. Where does the point
    𝐶

    𝑎

    𝑎

    C⋅
    ∣a∣
    a

    come from?
    This is the actual point on the vector
    𝑎
    a's line that is:

Distance
𝐶
C away in the direction of
𝑎
a (since
𝑎
a is a vector),

Scaled by the unit vector in that direction:
𝑎

𝑎

∣a∣
a

So:

Projection point

𝐶

𝑎

𝑎

Projection point=C⋅
∣a∣
a

This point is where all the projections of
𝑟
r's in the set fall. It's the foot of the perpendicular from the hyperplane onto the vector
𝑎
a.

📐 Diagram (conceptual)
pgsql
Copy code
*
/ | <-- vector r
/ |
/ |
/ | <-- All these r's project to the same spot
/ |
------------------------>
^ ^
origin projection point = C * (a / |a|)

           -->         direction of vector a

✅ Summary
The correct projection point is:

𝐶

𝑎

𝑎

C⋅
∣a∣
a

because it describes a fixed location on the line in direction of
𝑎
a.

The set of all vectors
𝑟
r with dot product
𝑟

𝑎

𝐶
r⋅a=C form a hyperplane orthogonal to
𝑎
a, and all of them project onto the same point.

Let me know if you'd like an actual drawn diagram or want to see this in code or coordinate form!

@mhayter mhayter closed this Aug 8, 2025
@adamant-pwn
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Substituting $r = C \frac{a}{|a|^2}$ and from $a \cdot a = |a|^2$, we see that, indeed, $r \cdot a = C\frac{|a|^2}{|a|^2}=C$, so it actually should be $|a|^2$.

@mhayter please don't take ChatGPT output for granted, it can often spill wrong information.

Note: ChatGPT 5 gives correct expression for it.

@adamant-pwn adamant-pwn reopened this Aug 9, 2025
@adamant-pwn adamant-pwn merged commit 6bc8214 into cp-algorithms:main Aug 9, 2025
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3 participants