Problem 49 - Advanced Engineering Mathematics

Question

In Problems, find, if possible, an equation of a plane that contains the given points. $$ (1,2,-1),(4,3,1),(7,4,3) $$

Step-by-Step Solution

Answer

The points are collinear; no unique plane exists.

Step 1: Find the Vectors

First, we need to find two vectors in the plane by using the given points. Let's label the points as \(A(1,2,-1)\), \(B(4,3,1)\), and \(C(7,4,3)\). Calculate the vectors: \( \vec{AB} = B - A = (4-1, 3-2, 1+1) = (3, 1, 2) \) and \( \vec{AC} = C - A = (7-1, 4-2, 3+1) = (6, 2, 4) \).

Step 2: Compute the Cross Product

The normal vector \( \vec{n} \) to the plane can be found using the cross product \( \vec{AB} \times \vec{AC} \). Use the determinant formula: \( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ 6 & 2 & 4 \end{vmatrix} = \hat{i}(1*4 - 2*2) - \hat{j}(3*4 - 2*6) + \hat{k}(3*2 - 1*6). \) Simplifying, \( \vec{n} = \hat{i}(4 - 4) - \hat{j}(12 - 12) + \hat{k}(6 - 6) = \textbf{0}. \)

Step 3: Determine If a Plane Exists

Since the cross product of the vectors results in the zero vector \(\textbf{0} = (0,0,0)\), this indicates that the points are collinear. A zero normal vector shows there is no unique plane (or infinitely many planes) that contain these points.

Step 4: Conclusion

Given that the vectors \( \vec{AB} \) and \( \vec{AC} \) are parallel (since they produce a zero cross-product), the points \(A, B, C\) are collinear. Therefore, it's not possible to define a unique plane using these points.

Key Concepts

Vectors in 3D Geometry

In 3D geometry, vectors represent quantities that have both a direction and a magnitude. When dealing with points in space, like in our problem, vectors are used to connect these points, forming segments. For instance, given three points

• \(A(1,2,-1)\)
• \(B(4,3,1)\)
• \(C(7,4,3)\)

Connecting these, the vectors \( \vec{AB} \) and \( \vec{AC} \) are formed by subtracting the coordinates of the starting point from the ending point. This operation is crucial in 3D space to establish relationships between points. Each vector can be visualized as an arrow starting from one point and ending at the other, fully described by its coordinates.

Cross Product

The cross product is a mathematical operation that takes two vectors in three-dimensional space and returns a new vector perpendicular to the plane containing the original vectors. It's an essential tool in determining the orientation of two vectors in 3D space. Given two vectors \( \vec{u} \) and \( \vec{v} \), their cross product, denoted as \( \vec{u} \times \vec{v} \), is calculated using the determinant method.
For vectors \( \vec{AB} = (3, 1, 2) \) and \( \vec{AC} = (6, 2, 4) \) from our exercise, the cross product formula helps determine the normal vector to the plane. However, if this cross product results in the zero vector, it indicates that our vectors do not span an area, suggesting collinearity or parallelism between them.

Collinearity

Collinearity refers to the condition where a set of points lie on a single straight line. In our exercise, when calculating the cross product of the vectors \( \vec{AB} \) and \( \vec{AC} \), we obtained a zero vector, \( \textbf{0} = (0,0,0) \). This zero normal vector is a strong indication that points \(A, B, C\) are collinear. If the cross product is zero, it often means that there is no plane that uniquely passes through the given set of points. Instead, these points simply lie on the same line. Therefore, it is only possible to discuss infinite planes parallel to this line, but not a unique one containing these points.

Normal Vector

A normal vector is a vector that is perpendicular to a surface or plane in a given space. It is a crucial component when defining the orientation of a plane. In the context of our exercise, the cross product \( \vec{AB} \times \vec{AC} \) is supposed to yield this normal vector. This vector defines the plane's direction in 3D space.
In a typical scenario where the points aren't collinear, the cross product results in a non-zero normal vector, allowing us to construct an equation for the plane. However, since we obtained a zero vector, it implies either a lack of a well-defined normal or multiple planes, thereby preventing us from determining a specific plane that passes through all the points simultaneously.