Write step-by-step solution for this problem:

Given the set of vectors A=Span{(2,0,1),(1,3,4)} and a vector b=(5,6,7), find the point in the span of A that is nearest to b.

To find the point in the span of \( A = \text{Span}\{(2,0,1),(1,3,4)\} \) that is nearest to \( \mathbf{b} = (5, 6, 7) \), we can use the concept of projection. Here's a step-by-step solution:

### Step 1: Set Up the Problem
We need to find the vector \(\mathbf{p}\) in the span of \( A \) such that \(\mathbf{p} \) is the closest point to \(\mathbf{b} \). 

Let the vectors in the span of \( A \) be \(\mathbf{v}_1 = (2, 0, 1)\) and \(\mathbf{v}_2 = (1, 3, 4)\). Any vector in the span of \( A \) can be written as:
\[
\mathbf{p} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = c_1 (2, 0, 1) + c_2 (1, 3, 4)
\]
\[
\mathbf{p} = (2c_1 + c_2, 3c_2, c_1 + 4c_2)
\]

### Step 2: Formulate the Problem as a System of Equations
We want to minimize the distance between \(\mathbf{p}\) and \(\mathbf{b}\). The distance (squared) between \(\mathbf{p}\) and \(\mathbf{b}\) is:
\[
\|\mathbf{p} - \mathbf{b}\|^2 = (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2
\]

### Step 3: Solve for \(c_1\) and \(c_2\)
To find the minimum, we take the partial derivatives of \(\|\mathbf{p} - \mathbf{b}\|^2\) with respect to \(c_1\) and \(c_2\), and set them to zero.

\[
\frac{\partial}{\partial c_1} \left[ (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2 \right] = 0
\]
\[
\frac{\partial}{\partial c_2} \left[ (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2 \right] = 0
\]

This results in the following system of linear equations:

1. \[
2(2c_1 + c_2 - 5) \cdot 2 + 2(c_1 + 4c_2 - 7) \cdot 1 = 0
\]
2. \[
2(2c_1 + c_2 - 5) \cdot 1 + 2(3c_2 - 6) \cdot 3 + 2(c_1 + 4c_2 - 7) \cdot 4 = 0
\]

### Step 4: Simplify the Equations
Simplify the equations:

1. \[
8c_1 + 4c_2 - 20 + 2c_1 + 8c_2 - 14 = 0
\]
\[
10c_1 + 12c_2 = 34
\]
\[
5c_1 + 6c_2 = 17 \quad \text{(Equation 1)}
\]

2. \[
4c_1 + 2c_2 - 10 + 18c_2 - 36 + 4c_1 + 16c_2 - 28 = 0
\]
\[
8c_1 + 36c_2 = 74
\]
\[
4c_1 + 18c_2 = 37 \quad \text{(Equation 2)}
\]

### Step 5: Solve the System of Equations
Solve the linear system:
\[
5c_1 + 6c_2 = 17
\]
\[
4c_1 + 18c_2 = 37
\]

From Equation 1:
\[
c_1 = \frac{

Learn how to find the point in the span of vectors A that is nearest to vector b with a detailed step-by-step solution. This problem involves key concepts in vector spaces, linear algebra, and projection. Suitable for undergraduate students studying linear algebra.

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