<Question>
The set of vectors $\mathbf{v}$ such that
\[\operatorname{proj}_{\begin{pmatrix} 2 \\ 1 \end{pmatrix}} \mathbf{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]lie on a line.  Enter the equation of this line in the form "$y = mx + b$".
</Question>

<think>
<Plan_1>
Okay, let's see. The problem is asking about the set of vectors v such that the projection of v onto the vector (2, 1) is the zero vector. And we need to find the equation of the line on which these vectors lie, in the form y = mx + b. Hmm. Alright, let me start by recalling what a projection is.

<prompt_1.1> Recall the formula for the projection of a vector onto another vector. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The projection of vector v onto another vector u is given by the formula:

proj_u v = ( (u · v) / ||u||² ) u

Right? So, this formula scales the vector u by the factor (u · v) divided by the squared norm of u. So, if the projection is the zero vector, that means that ( (u · v) / ||u||² ) u = 0. Since u is the vector (2, 1), which is not the zero vector, the scalar factor (u · v) / ||u||² must be zero. Because if the scalar is zero, then multiplying by u (which is non-zero) gives the zero vector. So, that implies that (u · v) / ||u||² = 0. Which further implies that u · v = 0. Because the denominator ||u||² is the squared norm of u, which is 2² + 1² = 5, so it's non-zero. Therefore, the condition simplifies to u · v = 0. Wait, but u is (2, 1), so u · v = 2*v1 + 1*v2 = 0. So, the set of vectors v = (v1, v2) such that 2v1 + v2 = 0. But the problem says that these vectors lie on a line, which makes sense because the equation 2v1 + v2 = 0 is a linear equation in two variables, so it's a line.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but let me check again. The projection being zero means that v is orthogonal to u?

<prompt_2.1> Consider the geometric interpretation of the projection being zero. </prompt_2.1>
-----

Let me see.

Alternatively, maybe I need to consider the projection formula again. Let me write it out step by step.

<prompt_2.2> Re-evaluate the problem using the projection formula and confirm the line equation. </prompt_2.2>
-----

But wait, let me check if there's another way to approach this. For instance, the set of vectors orthogonal to u = (2,1) is indeed a line, which is the orthogonal complement of the span of u. Since u is in R², its orthogonal complement is the line through the origin perpendicular to u. But in the problem, the projection being zero is equivalent to v being orthogonal to u, so yes, that line is 2x + y = 0, which is the same as y = -2x + 0.

<prompt_2.3> Explore alternative methods to confirm the line equation. </prompt_2.3>

But wait, let me just visualize this to make sure. If we have the vector (2,1), then the line y = -2x is perpendicular to it.

<prompt_2.4> Visualize the line and verify with specific examples. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Because if the projection is zero, that means v is orthogonal to u. So, the set of all vectors orthogonal to u is the orthogonal complement of the span of u, which in R² is a line through the origin. But the problem says "lie on a line". So, that line is the set of all vectors orthogonal to u = (2,1). Therefore, the equation should be 2x + y = 0, which can be written as y = -2x. So, the line is y = -2x.
</execution_2.1>

<execution_2.2>
Let u = (2, 1) and v = (v1, v2). The projection of v onto u is:

proj_u v = ( (u · v) / ||u||² ) u

We want this to be the zero vector. So:

( (u · v) / ||u||² ) u = 0

Since u is not the zero vector, the scalar (u · v) / ||u||² must be zero. Therefore, u · v = 0. So:

2*v1 + 1*v2 = 0

Which is a linear equation in v1 and v2. So, this is the equation of a line. To write it in the form y = mx + b, we can solve for v2 in terms of v1:

v2 = -2*v1

So, if we let x = v1 and y = v2, then the equation is y = -2x + 0. So, the line is y = -2x.
</execution_2.2>

<execution_2.3>
So, that's the line we need. Therefore, the answer should be y = -2x. But let me confirm this. Let's take a vector on this line, say (1, -2). Then the projection onto (2,1) should be zero. Let's compute it:

proj_(2,1) (1, -2) = ( (2*1 + 1*(-2)) / (2² + 1²) )*(2,1) = (2 - 2)/5*(2,1) = 0*(2,1) = (0,0). Correct. Similarly, take another point on the line, say (0,0). Well, (0,0) is on the line, and proj_(2,1) (0,0) is (0,0). But (0,0) is the zero vector, which is trivially the projection. So, that's okay. Wait, but the zero vector is on every line, but in this case, it's part of the solution set because the projection of zero is zero. So, that's correct.

Alternatively, if we take a vector not on this line, say (1, 0), then the projection onto (2,1) would be ( (2*1 + 1*0)/5 )*(2,1) = (2/5)*(2,1) = (4/5, 2/5), which is not zero. So, that vector is not in the solution set, which is consistent with it not being on the line y = -2x. Because if x=1, then y = -2(1) = -2, but this vector is (1,0), which is not on that line. So, the projection is non-zero, which is correct.

Therefore, all vectors v such that their projection onto (2,1) is zero are exactly those vectors that are orthogonal to (2,1), which form the line 2x + y = 0, or y = -2x. Therefore, the equation of the line is y = -2x + 0, so y = -2x.
</execution_2.3>

<execution_2.4>
Yes, because the slope of the line is -2, and the original vector (2,1) has a slope of 1/2, so they are perpendicular. So, that makes sense. For example, take the vector (-1, 2), which is on the line y = -2x. The projection of (-1, 2) onto (2,1) would be ( (2*(-1) + 1*2 ) / (2² + 1²) )*(2,1) = ( (-2 + 2)/5 )*(2,1) = 0*(2,1) = (0,0).
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Wait, but the problem says "enter the equation of this line in the form 'y = mx + b'".
So, in this case, the equation is y = -2x + 0. So, they might want it written as y = -2x, but let me check if there's a possibility I made a mistake in the sign.

<prompt_3.1> Re-evaluate the dot product and confirm the line equation. </prompt_3.1>
-----

Based on execution_2.3:
Therefore, the answer is y = -2x. But just to make sure, let's think again.

<prompt_3.2> Perform a final verification of the solution and conclude. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me verify the dot product again. If u = (2,1) and v = (v1, v2), then u · v = 2v1 + v2. So, setting that equal to zero gives 2v1 + v2 = 0, which can be rewritten as v2 = -2v1. So, if v1 is x, then y = -2x. So, the slope is -2, which is correct. So, the equation is y = -2x, which is the same as 2x + y = 0. So, yes, that's correct.
</execution_3.1>

<execution_3.2>
If we have a vector v such that proj_u v = 0, then v is orthogonal to u. So, the set of all such v is the orthogonal complement of the span of u, which is a line in R². The equation of that line is indeed 2x + y = 0, which is y = -2x. So, yes, that's correct. Therefore, the answer is y = -2x. I think that's solid. Let me check if there's another perspective.

Suppose we write the projection formula again. For the projection to be the zero vector, the formula ( (u · v) / ||u||² ) u = 0 must hold. Since u is non-zero, the scalar must be zero, so u · v = 0. So, that's the same as the dot product being zero, which is the definition of orthogonality. Therefore, the set of vectors v orthogonal to u form the solution set. So, that's the orthogonal complement, which is a line through the origin with slope -2. Therefore, the equation is y = -2x. So, I think that's correct. I don't see any mistakes here. So, the answer should be y = -2x.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
**Final Answer**
The equation of the line is \boxed{y = -2x}.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the equation of the line on which these vectors lie is $\boxed{y = -2x}$.
</Final_answer>
</think>