91Ó°ÊÓ

The cross product is an essential operation in vector calculus, especially when you're dealing with planes. It's a way to find a vector that is perpendicular to two given vectors. This perpendicular vector is called the "normal vector." In the context of finding the equation of a plane, the cross product is used to determine the normal vector from two vectors that lie within the plane.

To compute the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), use the determinant method:

• Set up a 3x3 matrix with the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row.
• Place your first vector components in the second row and your second vector components in the third row.
• Compute the determinant to find the components of the cross product.

In this exercise, we find the cross product of \( \mathbf{v_1} = \langle 0, -6, 4 \rangle \) and \( \mathbf{v_2} = \langle 0, -16, 6 \rangle \), which gives us the normal vector \( \mathbf{n} = \langle 64, 0, 0 \rangle \). This result is crucial for determining the plane's equation.

A normal vector is perpendicular to a plane and is fundamental for defining the plane's equation. When you have the normal vector, you can write the equation of the plane in the form:\( A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \), where \(\langle A, B, C \rangle\) is the normal vector to the plane.

For the plane through points \((1,9,2), (1,3,6), (1,-7,8)\), we use the cross product of two vectors lying in the plane to find the normal vector. Calculated as \( \langle 64, 0, 0 \rangle \), this vector indicates that the plane is vertical, oriented parallel to the x-axis.

Understanding the role of the normal vector helps us determine the "tilt" of the plane in three-dimensional space. If the normal vector has components only in one direction, like \( \langle 64, 0, 0 \rangle \), it suggests symmetry that often simplifies the plane's equation.

When dealing with a plane defined by three points, we can derive two vectors that lie within the plane by subtracting coordinates of these points. These vectors help us understand the plane's orientation and enable us to calculate the normal vector using the cross product.

In our exercise, the points \((1,9,2)\), \((1,3,6)\), and \((1,-7,8)\) provide us opportunities to create vectors:

• \( \mathbf{v_1} = (1,3,6) - (1,9,2) = \langle 0, -6, 4 \rangle \)
• \( \mathbf{v_2} = (1,-7,8) - (1,9,2) = \langle 0, -16, 6 \rangle \)

Both vectors are essential to form a complete understanding of the plane's structure. By subtracting the corresponding points' components, the vectors effectively "capture" the plane's internal dimensions. These vectors are then used to find the cross product, ultimately leading to the equation of the plane.

To determine the equation of a plane passing through three points, we first need to find a normal vector using the cross product of vectors in the plane. The plane's equation is then derived using these points and the normal vector.

Given points in this problem are \((1,9,2)\), \((1,3,6)\), and \((1,-7,8)\). Through these coordinates, vectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \) provide us the necessary components to calculate the normal vector \( \mathbf{n} = \langle 64, 0, 0 \rangle \).

Using one of the points, say \( (1, 9, 2) \), and the normal vector \( \langle 64, 0, 0 \rangle \), we can write the plane's equation. The equation simplifies to \( x = 1 \), indicating a plane that is completely vertical at \( x = 1 \).

This approach underpins many problems involving planes in geometry, showing how cross products and vectors can simplify the task of finding plane equations through specific points.