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We are given two vectors, \( \vec{A} = A_x \hat{i} + 7 \hat{j} + 2 \hat{k} \) and \( \vec{B} = B_x \hat{i} + 2 \hat{j} + 1 \hat{k} \), along with the information that:

1. \( \vec{A} - \vec{B} = 2 \hat{i} + 5 \hat{j} + 1 \hat{k} \)
2. The magnitude (modulus) of \( \vec{A} \) is 10.

### Step 1: Solve for \( A_x \) and \( B_x \)

Using the first piece of information \( \vec{A} - \vec{B} = 2 \hat{i} + 5 \hat{j} + 1 \hat{k} \), we know:

\[
(A_x - B_x) \hat{i} + (7 - 2) \hat{j} + (2 - 1) \hat{k} = 2 \hat{i} + 5 \hat{j} + 1 \hat{k}
\]

From this, we can compare components:

- In the \( \hat{i} \)-direction: \( A_x - B_x = 2 \)
- In the \( \hat{j} \)-direction: \( 7 - 2 = 5 \), which is consistent.
- In the \( \hat{k} \)-direction: \( 2 - 1 = 1 \), which is also consistent.

So, we now have the equation:

\[
A_x - B_x = 2 \tag{1}
\]

### Step 2: Use the modulus of \( \vec{A} \)

We are also told that the magnitude of \( \vec{A} \) is 10, so:

\[
|\vec{A}| = \sqrt{A_x^2 + 7^2 + 2^2} = 10
\]

This simplifies to:

\[
\sqrt{A_x^2 + 49 + 4} = 10
\]
\[
\sqrt{A_x^2 + 53} = 10
\]
\[
A_x^2 + 53 = 100
\]
\[
A_x^2 = 47
\]
\[
A_x = \pm \sqrt{47}
\]

Thus, \( A_x = \sqrt{47} \) or \( A_x = -\sqrt{47} \).

### Step 3: Solve for \( B_x \)

From equation (1), we know \( A_x - B_x = 2 \). So:

\[
B_x = A_x - 2
\]

Substituting the two possible values for \( A_x \):

- If \( A_x = \sqrt{47} \), then \( B_x = \sqrt{47} - 2 \).
- If \( A_x = -\sqrt{47} \), then \( B_x = -\sqrt{47} - 2 \).

### Step 4: Find the angle that \( \vec{A} \) forms with the x-axis

The angle \( \theta \) that \( \vec{A} \) forms with the x-axis is given by:

\[
\cos \theta = \frac{A_x}{|\vec{A}|}
\]

Since \( |\vec{A}| = 10 \) and \( A_x = \pm \sqrt{47} \), we have:

\[
\cos \theta = \frac{\pm \sqrt{47}}{10}
\]
\[
\theta = \cos^{-1} \left( \frac{\pm \sqrt{47}}{10} \right)
\]

Thus, the angle \( \theta \) is \( \cos^{-1} \left( \frac{\sqrt{47}}{10} \right) \) or \( \cos^{-1} \left( \frac{-\sqrt{47}}{10} \right) \).

---

Would you like more details on any step, or do you have questions?

Here are some related questions to further your understanding:

1. How is the modulus of a vector calculated in 3D?
2. What is the geometric meaning of the dot product in relation to angles?
3. How do different components of a vector contribute to its overall magnitude?
4. What are some methods to visualize vectors in three-dimensional space?
5. How do angles between vectors and coordinate axes help in physics and engineering problems?

**Tip**: When finding the angle between a vector and an axis, the component along that axis divided by the vector's magnitude gives the cosine of the angle.

Dado los vectores A⃗ = Axi⃗ + 7j⃗ + 2k⃗ y B⃗ = Bxi⃗ + 2j⃗ + 1k⃗, sabiendo que A⃗ − B⃗ = 2i⃗ + 5j⃗ + 1k⃗ y el módulo de la suma vale 10. Determina lo siguiente: Los valores de Ax y Bx y el ángulo que forma el vector A⃗ con el eje de coordenadas.

Solve for the components Ax and Bx of two vectors and determine the angle that vector A forms with the coordinate axis. Includes vector magnitude calculations and trigonometry for angle determination.

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