Problem 17 - Calculus Early Transcendentals

Question

Find an equation of the following planes. The plane passing through the points \((1,0,3),(0,4,2),\) and (1,1,1)

Step-by-Step Solution

Answer

Question: Find the equation of the plane passing through the points A(1,0,3), B(0,4,2), and C(1,1,1). Answer: The equation of the plane passing through the given points is \(x+y+z=5\).

Step 1: Find direction vectors

First, we need to find two direction vectors connecting the given points. We can do this by subtracting coordinates of these points. Let's denote the points as \(A(1,0,3)\), \(B(0,4,2)\), and \(C(1,1,1)\). We can find the direction vectors \(AB\) and \(AC\) as follows: \(\vec{AB} = B - A = (0-1, 4-0, 2-3) = (-1, 4, -1)\) \(\vec{AC} = C - A = (1-1, 1-0, 1-3) = (0, 1, -2)\)

Step 2: Find the normal vector

Now, we will find the normal vector to the plane by taking the cross product of direction vectors \(\vec{AB}\) and \(\vec{AC}\): \(\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} i & j & k \\ -1 & 4 & -1 \\ 0 & 1 & -2 \end{vmatrix} = \ (-3-(-4))i - (-1+0)j - (1-0)k = (1,1,1)\) The normal vector to the plane is \(\vec{n} = (1,1,1)\).

Step 3: Find the equation of the plane

With the normal vector and a point on the plane (let's use point A), we can now plug these values into the general equation of a plane: \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\) Here, \(A\), \(B\), and \(C\) are the components of the normal vector, and \((x_0, y_0, z_0)\) are the coordinates of point A. Substituting the values, we get: \((1)(x-1) + (1)(y-0) + (1)(z-3) = 0\) After simplification, we get the equation of the plane as: \(x + y + z = 5\) So, the equation of the plane passing through the given points is \(x+y+z=5\).

Key Concepts

Cross Product

The cross product is an essential operation in vector algebra that helps us find a vector perpendicular to two given vectors in three-dimensional space. This is important in the case of planes, as finding a perpendicular vector (or normal vector) can help us form the equation of a plane. To find the cross product, you'll often see a determinant involving the unit vectors: \(i, j, \) and \(k\), placed alongside the components of the given vectors.

• The result of a cross product is always a vector.
• It represents the area of the parallelogram formed by the two vectors.
• The direction follows the right-hand rule, usually perpendicular to the plane formed by the original vectors.

In the step-by-step solution, the cross product of vectors \(\vec{AB}\) and \(\vec{AC}\) results in \(\vec{n} = (1, 1, 1)\), which is the normal vector to the plane.

Direction Vectors

Direction vectors play a key role in defining the geometry of planes. When given points through which a plane passes, direction vectors are derived by connecting these points through vectors. This involves subtracting the coordinates of the starting point from the endpoint to form a new vector.

• Direction vectors help describe the orientation of a plane.
• They are used to construct the normal vector through a cross product.

In the original exercise solution, the direction vectors \(\vec{AB} = (-1, 4, -1)\) and \(\vec{AC} = (0, 1, -2)\) were found by such subtraction, illustrating how changes in coordinates can guide us in visualizing the orientation of the plane.

Normal Vector

A normal vector to a plane is fundamental because it is perpendicular to the entire plane. This concept is central to the equation of the plane. By utilizing the cross product of two direction vectors lying on the plane, we can find this normal vector. The components of this vector are integrated into the equation of the plane.

• The normal vector provides the coefficients \(A, B,\) and \(C\) in the plane equation \(Ax + By + Cz = D\).
• Its perpendicularity serves as a basis for many geometric computations.

In the step-by-step guide, the normal vector \(\vec{n} = (1, 1, 1)\) was derived, forming the basis for the final equation of the plane.

Linear Algebra

Linear algebra provides the tools necessary to work with vectors and planes in multidimensional spaces. It involves operations and equations that help us define and manipulate lines, planes, and other hyperplanes. Understanding basic operations like vector addition, subtraction, and particularly the cross product are crucial in constructing geometric forms such as planes.

• It emphasizes the study of vectors, vector spaces, and linear transformations.
• Linear algebra concepts enable us to find intersections, compute areas, and solve equations of planes.

In solving the original exercise, linear algebra techniques are used to derive direction vectors, thereby obtaining the normal vector, and finally, formulating the equation of the plane \(x + y + z = 5\). This process underscores the importance of linear algebra in understanding multidimensional geometry.