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Understanding the cross product is essential when dealing with vectors in 3-dimensional space, particularly when trying to find a vector that is perpendicular to two given vectors. In the context of planes in three dimensions, the cross product helps us to determine a normal vector to the plane.

Here's how the cross product is computed: for two vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), their cross product \( \vec{c} = \vec{a} \times \vec{b} \) is a vector given by \( \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).

This resulting vector \( \vec{c} \) is orthogonal to both \( \vec{a} \) and \( \vec{b} \). To clarify, being orthogonal means that \( \vec{c} \) is at a 90-degree angle to both \( \vec{a} \) and \( \vec{b} \). This is crucial for establishing the orientation of a plane in 3D space as it determines the plane's 'up' direction. In the case of the textbook exercise, the cross product was used to find a vector that is perpendicular to the two given planes.

A normal vector is a fundamental component in the equation of a plane as it represents a vector that is perpendicular to the surface of the plane. Whenever we talk about the equation of a plane, the normal vector essentially dictates the orientation of the plane in three-dimensional space.

A plane is defined mathematically by its normal vector \( \vec{n} = \langle n_x, n_y, n_z \rangle \) and a point through which the plane passes \( (x_0, y_0, z_0) \) by the equation \( n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \). Here's a closer look at the components:

• \(\vec{n}\): The normal vector.
• \( (x_0, y_0, z_0) \): A given point on the plane.
• \( (x, y, z) \): Any point on the plane.

Every point \( (x, y, z) \) that satisfies this equation lies on the plane, and the normal vector is key to defining its properties. In the exercise, after finding the cross product of the normal vectors of the two given planes, the resultant vector \( \vec{n} \) serves as the normal vector for the equation of the plane passing through the point \( (0, -2, 4) \).

Orthogonal planes are planes that intersect each other at a right angle. In a three-dimensional Cartesian coordinate system, if two planes are orthogonal, their normal vectors are also orthogonal to each other.

To determine if two planes are orthogonal, one can check if the dot product of their normal vectors is zero, because dot product is a measure of how parallel or perpendicular two vectors are. If the dot product is zero, the vectors—and hence the planes they represent—are orthogonal.

In practical scenarios, such as the one depicted in the textbook exercise, we are sometimes interested in finding a third plane that is orthogonal to two given planes. This is achieved through the cross product of the normal vectors of the given planes, yielding a normal vector for the sought orthogonal plane. The resulting normal vector is perpendicular to both of the original planes' normal vectors, ensuring that the resulting plane is orthogonal to the given ones. The exercise walks through finding a plane that is orthogonal to the two given planes by first determining the cross product of their normal vectors, then using this new normal vector to write the plane's equation.