91Ó°ÊÓ

To understand this exercise, let's first dive into the concept of orthogonal vectors. Two vectors are said to be orthogonal if their dot product equals zero. In mathematical terms, if \( \mathbf{a} \cdot \mathbf{b} = 0 \), then vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.

In this problem, the cross product \( \mathbf{a} \times \mathbf{x} = \mathbf{b} \) results in a vector \( \mathbf{b} \) that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{x} \). This orthogonality is key to understanding why \( \mathbf{a} \cdot \mathbf{b} = 0 \), because every result of a cross product is by definition orthogonal to its operand vectors.

When you see \( \mathbf{a} \cdot \mathbf{b} = 0 \) confirmed in Step 2 of the solution, this orthogonality property of the cross product is being utilized. Essentially, this is what allows us to state that the vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular in three-dimensional space.

Another crucial concept in this exercise is the vector triple product. This is a specific identity in vector algebra given by:
\[ \text{If } \mathbf{u}, \mathbf{v}, \text{ and } \mathbf{w} \text{ are vectors, then } \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}. \]
In Step 6 of the solution, the identity is applied to simplify the term \( \mathbf{a} \times (\mathbf{b} \times \mathbf{a}) \).
By plugging vectors into this identity, we determine:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{a}) = (\mathbf{a} \cdot \mathbf{a})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{a}. \]
Here, we know from earlier that \( \mathbf{a} \cdot \mathbf{b} = 0 \). So the equation simplifies to:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{a}) = ||\mathbf{a}||^{2}\mathbf{b}. \]
This identity helps us handle more complex expressions involving cross products by breaking them down into simpler dot products and scalar multiplications.

In this exercise, we are given a vector equation \( \mathbf{a} \times \mathbf{x} = \mathbf{b} \) and asked to find a solution for the vector \( \mathbf{x} \). The proposed solution is:
\[ \mathbf{x} = \frac{\mathbf{b} \times \mathbf{a}}{||\mathbf{a}||^{2}} + k \mathbf{a}, \]
where \( k \) is any scalar. This solution utilizes the properties of the cross product and vector identities (namely, the vector triple product).

Here’s a step-by-step rundown:

• First, substitute the proposed solution into the original vector equation: \( \mathbf{a} \times \left( \frac{\mathbf{b} \times \mathbf{a}}{||\mathbf{a}||^{2}} + k \mathbf{a} \right) = \mathbf{a} \times \frac{\mathbf{b} \times \mathbf{a}}{||\mathbf{a}||^{2}} + \mathbf{a} \times (k \mathbf{a}) \).


• Next, evaluate each term separately. For \( \mathbf{a} \times (k \mathbf{a}) \), it simplifies to \( \mathbf{0} \), because a vector crossed with itself is zero.


• Then, apply the vector triple product identity to the term \( \mathbf{a} \times \left( \frac{\mathbf{b} \times \mathbf{a}}{||\mathbf{a}||^{2}} \right) \), leading to the simplification: \( \frac{\| \mathbf{a} \|^{2}\mathbf{b}}{||\mathbf{a}||^{2}} = \mathbf{b} \).


• Finally, combining all parts, we get \( \mathbf{b} = \mathbf{b} \), which confirms that the solution holds true.

This means \( \mathbf{x} = \frac{\mathbf{b} \times \mathbf{a}}{||\mathbf{a}||^{2}} + k \mathbf{a} \) is indeed a valid solution to our original vector equation. The presence of \( k \mathbf{a} \) indicates that there is an infinite number of solutions, each differing by a vector parallel to \( \mathbf{a} \).