91Ó°ÊÓ

The cross product is a mathematical operation used when dealing with vectors in three-dimensional space. It results in a new vector that is perpendicular to the plane formed by the original two vectors. To calculate the cross product of two vectors, we utilize a determinant and the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). In the context of finding the line of intersection between two planes, we use the cross product to find a direction vector of the resulting line.
For example, in the problem we have two normal vectors: \( \mathbf{n_1} = (2, 3, 1) \) and \( \mathbf{n_2} = (1, 3, 2) \). The cross product is calculated as:

• The determinant is represented using the unit vectors.
• Each component of the resulting vector is found as a minor determinant, creating the vector \( \mathbf{d} = (3, -3, 3) \).

This cross product gives us the necessary direction vector that is aligned with the line of intersection.

The direction vector is a crucial component when determining the line where two planes intersect. It reveals the path along which the line extends in three-dimensional space. When the two planes intersect, the line formed has a direction influenced by both plane's normal vectors.
The direction vector \( \mathbf{d} \) can be acquired by calculating the cross product of the normal vectors, as shown above. This provides a vector that is perpendicular to both normals, hence lying in the correct direction of the intersection line. The calculated direction vector here is \( (3, -3, 3) \).
This vector intimates the general orientation of the line, giving insight into the line's angle and movement in space.

Normalizing a vector involves scaling it so that it has a magnitude of 1, keeping its direction unchanged. This process simplifies many computations, like determining angles between vectors and their components. To normalize a vector, divide each of its components by its magnitude.
For the direction vector \( \mathbf{d} = (3, -3, 3) \), calculate the magnitude as \( 3\sqrt{3} \). Once normalized, it becomes:\(\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)
This vector preserves the line's directionality while rendering further mathematical operations more straightforward due to its unit magnitude.

When a line in space is represented by a direction vector, finding the angle it makes with any axis requires calculating the cosine of that angle. Cosine of the angle \( \alpha \) with an axis is determined by

• Taking the dot product of the normalized direction vector with the unit vector of the desired axis (e.g., \( \mathbf{i} = (1, 0, 0) \) for the x-axis).
• The result of the dot product gives the cosine of the angle with respect to that axis.

In our scenario, to find the angle \( \alpha \)'s cosine that the line makes with the x-axis, compute the dot product of the normalized direction vector \( \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) and the x-axis unit vector \( (1, 0, 0) \), which results in \( \frac{1}{\sqrt{3}} \). This calculation shows how the line aligns in terms of angle with the x-axis, aiding in understanding the line's spatial orientation.