91Ó°ÊÓ

To understand the concept of vectors in a plane, visualize them as arrows. These arrows represent direction and magnitude. In the context of our exercise, we're interested in vector representation between points in three-dimensional space.
When dealing with planes, at least two vectors are required to delineate the plane's orientation. In the solution provided, vectors \(\vec{AB}\) and \(\vec{AC}\) were determined. They are obtained by calculating the difference in coordinates between the respective points.

• For \(\vec{AB}\), subtract coordinates of point A from B: \((1-0, 0-1, 1-1) = (1, -1, 0)\).
• For \(\vec{AC}\), subtract coordinates of point A from C: \((1-0, 1-1, 0-1) = (1, 0, -1)\).

These vectors essentially lie within the desired plane and will guide us to find the unique plane that passes through these points.

The cross product is a valuable vector operation in three dimensions. It helps us find a vector that is perpendicular to the plane defined by two vectors. You take two vectors, say \(\vec{A}\) and \(\vec{B}\), and calculate their cross product, \(\vec{A} \times \vec{B}\).
For vectors \(\vec{AB}\) and \(\vec{AC}\) identified in the earlier step, we apply the cross product operation to obtain a new vector that stands perpendicular to both, which is the normal vector to the plane.
The determination of \(\vec{AB} \times \vec{AC}\) involves a determinant calculation, which will be discussed in an upcoming section. Remember, cross products only apply in three-dimensional spaces and can be used to find the normal vector needed to define a plane.

A normal vector is crucial for establishing the orientation of a plane. It is a vector that is perpendicular to every vector lying in that plane. In our problem, after computing the cross product of \(\vec{AB}\) and \(\vec{AC}\), the resultant vector was \(\vec{n} = (1, 1, 1)\).
This vector \(\vec{n}\), being normal to the plane, directly gives the coefficients \((a, b, c)\) in the equation of the plane.
A normal vector is indispensable when you derive equations of planes, and it acts as an anchor, fixing the plane's tilt in the space.

The determinant is a special number that can be calculated from a square matrix. For our task, we use it as part of computing the cross product when determining the normal vector. When finding the cross product \(\vec{AB} \times \vec{AC},\) we form a 3x3 matrix with unit vectors \(\hat{i}, \hat{j}, \hat{k},\) and the components of \(\vec{AB}\) and \(\vec{AC}\).
The determinant is calculated to extract a vector from this setup. The calculation leads us to \(\vec{n} = (0, 1, 1)\), which simplifies to \((1, 1, 1)\). The determinant helps express relationships and properties in a matrix, crucial in many aspects of linear algebra, especially when computing vector operations like the cross product.

The equation of a plane combines all the components discussed above: the normal vector and a point on the plane. The general form is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where \((a, b, c)\) are coefficients from the normal vector, and \((x_0, y_0, z_0)\) is any of the points on the plane.
By choosing point A: \((0, 1, 1)\) and substituting into the equation with the normal vector \((1, 1, 1)\), we find the equation \(1(x - 0) + 1(y - 1) + 1(z - 1) = 0\).
Upon simplification, the equation becomes \(x + y + z = 2\), representing the specific plane that passes through the given points. This formula is a common method to describe the expanse and position of a plane in 3D space.