<Question>
If $\mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$ then find the vector $\mathbf{v}$ such that $\mathbf{a} \cdot \mathbf{v} = 2$ and $\mathbf{a} \times \mathbf{v} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$
</Question>

<think>
<Plan_1>
Okay, so I need to find a vector **v** such that when I take the dot product of **a** and **v**, I get 2, and when I take the cross product of **a** and **v**, I get the vector [1, -2, 1]. The vector **a** is given as [1, 1, 1]. Hmm, let me think about how to approach this.

First, I remember that the dot product and cross product are two different operations involving vectors. The dot product results in a scalar, and the cross product results in a vector that's perpendicular to both original vectors. Here, we have both conditions: the dot product equals 2, and the cross product equals a specific vector.
So, I need to find a vector **v** = [v1, v2, v3] that satisfies both these conditions.

Let me write down the given equations. The dot product equation is:

**a** ⋅ **v** = 1*v1 + 1*v2 + 1*v3 = v1 + v2 + v3 = 2.

The cross product **a** × **v** should equal [1, -2, 1]. Let me recall the cross product formula. If I have two vectors **a** = [a1, a2, a3] and **v** = [v1, v2, v3], then their cross product is:

<prompt_1.1> Write down the given equations for the dot product and cross product. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
[ (a2v3 - a3v2), (a3v1 - a1v3), (a1v2 - a2v1) ]

Given that **a** = [1, 1, 1], so substituting into the cross product formula:

First component: (1*v3 - 1*v2) = v3 - v2

Second component: (1*v1 - 1*v3) = v1 - v3

Third component: (1*v2 - 1*v1) = v2 - v1

And this cross product is supposed to equal [1, -2, 1]. So, setting up equations:

1. v3 - v2 = 1
2. v1 - v3 = -2
3. v2 - v1 = 1
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, let me check that cross product formula again to make sure I didn't mix up any components. The cross product **a** × **v** is calculated as:

<prompt_2.1> Recheck the cross product calculations for accuracy. </prompt_2.1>

So we have three equations:

1. v1 + v2 + v3 = 2 (from the dot product)
2. v3 - v2 = 1 (from the first component of the cross product)
3. v1 - v3 = -2 (from the second component of the cross product)
4. v2 - v1 = 1 (from the third component of the cross product)

Wait, equations 3 and 4 seem related. Let me check that.
Equation 3: v1 - v3 = -2

Equation 4: v2 - v1 = 1

So, if I add equations 3 and 4, I get:

(v1 - v3) + (v2 - v1) = -2 + 1 ⇒ v2 - v3 = -1

But from equation 2, we have v3 - v2 = 1, which is the same as -(v2 - v3) = 1 ⇒ v2 - v3 = -1.

So that's consistent.

So, we have a system of three equations:

1. v1 + v2 + v3 = 2
2. v3 - v2 = 1
3. v1 - v3 = -2

Let me write them again:

Equation 1: v1 + v2 + v3 = 2

Equation 2: -v2 + v3 = 1

Equation 3: v1 - v3 = -2

So, three equations with three variables. Let's solve this system.

<prompt_2.2> Solve the system of equations to find the vector **v**. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
i component: a2v3 - a3v2 = 1*v3 - 1*v2 = v3 - v2

j component: a3v1 - a1v3 = 1*v1 - 1*v3 = v1 - v3. But wait, in the cross product formula, the j component is negative of that, right? Because the cross product formula is:

**a** × **v** = (a2v3 - a3v2, a3v1 - a1v3, a1v2 - a2v1)

Wait, no, actually, the cross product formula in coordinates is:

If **a** = (a1, a2, a3) and **v** = (v1, v2, v3), then:

**a** × **v** = (a2v3 - a3v2, a3v1 - a1v3, a1v2 - a2v1)

Yes, so the components are:

First component: a2v3 - a3v2

Second component: a3v1 - a1v3

Third component: a1v2 - a2v1

So, substituting **a** = (1,1,1):

First component: 1*v3 - 1*v2 = v3 - v2

Second component: 1*v1 - 1*v3 = v1 - v3

Third component: 1*v2 - 1*v1 = v2 - v1

Therefore, the cross product **a** × **v** = (v3 - v2, v1 - v3, v2 - v1)

And this is supposed to equal (1, -2, 1). Therefore, the three equations are:

1. v3 - v2 = 1

2. v1 - v3 = -2

3. v2 - v1 = 1
</execution_2.1>

<execution_2.2>
From equation 3: v1 = v3 - 2

From equation 2: v3 = v2 + 1

So substitute v3 from equation 2 into equation 3: v1 = (v2 + 1) - 2 = v2 - 1

Now, substitute v1 and v3 in terms of v2 into equation 1.

Equation 1: (v2 - 1) + v2 + (v2 + 1) = 2

Simplify:

v2 - 1 + v2 + v2 + 1 = 2

Combine like terms:

3v2 = 2

Therefore, v2 = 2/3

Now, substitute back into equation 2: v3 = v2 + 1 = (2/3) + 1 = 5/3

And v1 = v2 - 1 = (2/3) - 1 = -1/3

So, the vector **v** is [ -1/3, 2/3, 5/3 ]
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Let me check if this satisfies all the equations.

<prompt_3.1> Verify the solution by checking if the vector **v** satisfies all the equations. </prompt_3.1>

Now, check the dot product: v1 + v2 + v3 = (-1/3) + (2/3) + (5/3) = ( (-1) + 2 + 5 ) / 3 = 6/3 = 2.

So, that's correct.

<prompt_3.2> Confirm the solution by checking both the dot product and cross product conditions. </prompt_3.2>

Wait, but let me think again. Is this the only solution? Since we had three equations with three variables, and we found a unique solution, that should be the only vector that satisfies both conditions. So, yes, this is the answer.

But just to make sure, let me try plugging the vector into the cross product formula again.

<prompt_3.3> Perform a final verification of the solution by recalculating the cross product. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
First, check equation 2: v3 - v2 = 5/3 - 2/3 = 3/3 = 1. Correct.

Equation 3: v1 - v3 = (-1/3) - (5/3) = (-6/3) = -2. Correct.

Equation 4: v2 - v1 = (2/3) - (-1/3) = 3/3 = 1. Correct.
</execution_3.1>

<execution_3.2>
Now, check the cross product:

**a** × **v** = (v3 - v2, v1 - v3, v2 - v1)

Compute each component:

First component: v3 - v2 = 5/3 - 2/3 = 3/3 = 1 ✔️

Second component: v1 - v3 = (-1/3) - 5/3 = (-6)/3 = -2 ✔️

Third component: v2 - v1 = (2/3) - (-1/3) = (2/3 + 1/3) = 3/3 = 1 ✔️

All components match the given cross product. So, the solution is correct.

Therefore, the vector **v** is [ -1/3, 2/3, 5/3 ]
</execution_3.2>

<execution_3.3>
Given **v** = [ -1/3, 2/3, 5/3 ]

Compute **a** × **v**:

First component: v3 - v2 = 5/3 - 2/3 = 3/3 = 1. Correct.

Second component: v1 - v3 = (-1/3) - 5/3 = (-6)/3 = -2. Correct.

Third component: v2 - v1 = (2/3) - (-1/3) = (2/3 + 1/3) = 3/3 = 1. Correct.

All components match the given cross product. And the dot product is 2. So, seems like this is the right answer.

But let me think again. The problem didn't specify any constraints on **v** other than the dot and cross product. Since **a** is a vector in 3D space, and we're given both a dot and cross product, there should be a unique solution. Because the cross product gives two equations (since cross product in 3D is a vector, so each component gives an equation), and the dot product gives another equation. So three equations for three variables, which we solved. So the solution is unique. Therefore, this is the only vector **v** that satisfies both conditions. So, I think this is the answer.
</execution_3.3>
</Execution_3>

<Final_answer>
Therefore, the vector $\mathbf{v}$ that satisfies the given conditions is $\boxed{\begin{pmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{5}{3} \end{pmatrix}}$.
</Final_answer>
</think>