91Ó°ÊÓ

The cross product is a fundamental operation in vector calculus that results in a vector perpendicular to two given vectors. If you have two vectors, say \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), then their cross product \( \mathbf{a} \times \mathbf{b} \) is calculated using the formula:\[\mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle.\]The result is not just any vector but one that stands perpendicular to both vectors involved in the operation.

• The magnitude of the cross product is equivalent to the area of the parallelogram formed by these vectors.
• The direction of the resulting vector follows the right-hand rule, which is a method to determine the orientation of the vector.

In our specific example, using vectors \( \langle 1, 1, 1 \rangle \) and \( \mathbf{w} \), we determined their cross product to form another vector \( \langle -1, -1, 2 \rangle \). This formulation aids in finding vectors that satisfy the vector equation.

Once we have the result from our cross product, we create a system of equations to solve for the unknown components of our vector \( \mathbf{w} \). Each element of the resultant vector, \( \langle -1, -1, 2 \rangle \), corresponds to an equation:

• \( w_3 - w_2 = -1 \)
• \( w_1 - w_3 = -1 \)
• \( w_2 - w_1 = 2 \)

This system of equations stems from equating the cross product result to the given vector. Such systems are valuable in linear algebra as they allow us to solve multiple equations simultaneously to find unknown variables.
The solution involves expressing variables in terms of one another, simplifying the system to find consistent or unique solutions.
In this problem, we discovered that the equations lead to a free variable, confirming the system's consistency and showing an infinite number of solutions lying in a particular direction.

Vector solutions in our context refer to the solutions we obtain for \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \) based on our system of equations. Here, we treated \( w_2 \) as a free variable - allowing us to express the other variables in terms of \( w_2 \). We derived:

• \( w_1 = w_2 - 2 \)
• \( w_3 = w_2 - 1 \)

As a consequence, any vector \( \mathbf{w} \) satisfying the original equation can be expressed as\(\mathbf{w} = \langle w_2 - 2, w_2, w_2 - 1 \rangle.\)The parameter \( w_2 \) represents a line of vectors, illustrating a direction in the vector space.
This notion of vector solutions is significant in understanding vector spaces since it provides insights into how solutions form patterns like lines or planes, each described by their dimensional freedom.