Let a = i + j - k and c = 2i - 3j + 2k. Then the number of vectors b such that b x c = a and |b|E{1, 2, …, 10} is …


Given the vectors \( \mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k} \) and \( \mathbf{c} = 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \), we need to find the number of vectors \( \mathbf{b} \) such that \( \mathbf{b} \times \mathbf{c} = \mathbf{a} \) and \( |\mathbf{b}| \in \{1, 2, \ldots, 10\} \).

### Step 1: Compute the cross product \( \mathbf{b} \times \mathbf{c} \)

Let \( \mathbf{b} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \). The cross product \( \mathbf{b} \times \mathbf{c} \) is given by the determinant:
\[
\mathbf{b} \times \mathbf{c} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x & y & z \\
2 & -3 & 2
\end{vmatrix}
\]
This expands to:
\[
\mathbf{b} \times \mathbf{c} = \mathbf{i} \left( y \cdot 2 - z \cdot (-3) \right) - \mathbf{j} \left( x \cdot 2 - z \cdot 2 \right) + \mathbf{k} \left( x \cdot (-3) - y \cdot 2 \right)
\]
Simplifying:
\[
\mathbf{b} \times \mathbf{c} = \mathbf{i} (2y + 3z) - \mathbf{j} (2x - 2z) + \mathbf{k} (-3x - 2y)
\]

### Step 2: Set \( \mathbf{b} \times \mathbf{c} = \mathbf{a} \)
We equate this to \( \mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k} \), leading to the system of equations:
\[
2y + 3z = 1 \quad \text{(coefficient of } \mathbf{i})
\]
\[
2z - 2x = 1 \quad \text{(coefficient of } \mathbf{j})
\]
\[
-3x - 2y = -1 \quad \text{(coefficient of } \mathbf{k})
\]

### Step 3: Solve the system of equations
Let's solve these equations step by step:

1. **From the first equation:**
   \[
   2y + 3z = 1
   \]

2. **From the second equation:**
   \[
   2z - 2x = 1 \quad \Rightarrow \quad z - x = \frac{1}{2} \quad \Rightarrow \quad z = x + \frac{1}{2}
   \]
   
3. **Substitute \( z = x + \frac{1}{2} \) into the first equation:**
   \[
   2y + 3\left(x + \frac{1}{2}\right) = 1 \quad \Rightarrow \quad 2y + 3x + \frac{3}{2} = 1 \quad \Rightarrow \quad 2y + 3x = -\frac{1}{2} \quad \Rightarrow \quad 4y + 6x = -1
   \]
   
4. **Using the third equation \( -3x - 2y = -1 \), multiply by 2:**
   \[
   -6x - 4y = -2
   \]
   
5. **Now, add the two equations:**
   \[
   (-6x - 4y) + (6x + 4y) = -2 - 1 \quad \Rightarrow \quad 0 = -3
   \]
   
This implies there is an inconsistency, indicating that the original equations were set up correctly for finding a relationship among \( x \), \( y \), and \( z \). The inconsistency suggests there's no solution for \( \mathbf{b} \) directly, so let's re-evaluate the process or check constraints.

However, considering the modulus condition \( |\mathbf{b}| \), and knowing that no direct solution exists, we must focus on alternative considerations, such as projections onto the null space.

To summarize the key result:
\[
\mathbf{b} \text{ vectors } \mathbf{b} \text{ satisfying } \mathbf{b} \times \mathbf{c} = \mathbf{a} \text{ need to fulfill the length condition.}
\]
There are no non-trivial vectors \( \mathbf{b} \) that satisfy all conditions simultaneously within the integer length space given by \( |\mathbf{b}| \).

Thus, the number of such vectors is **0**.

---

**Questions for Exploration:**
1. What is the general method for solving vector cross-product equations?
2. How does the modulus constraint affect the solution space?
3. Can you define orthogonal vectors and how they relate to cross products?
4. What happens when two vectors are parallel in cross products?
5. How does the determinant play a role in vector cross products?

**Tip:** Always check for consistency in your vector equations; inconsistencies may indicate no valid solutions exist for given constraints.

Let a = i + j - k and c = 2i - 3j + 2k. Then the number of vectors b such that b x c = a and |b|E{1, 2, …, 10} is …

Explore how to find the number of vectors b such that b × c = a, with specific magnitude constraints. This problem delves into vector algebra, including cross product computation and solving linear equations. Suitable for students in Grades 11-12.

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