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Chapter 13: Problem 17

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

Short Answer

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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 by step solution

01

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: \(\vec{AB} = B - A = (0-1, 4-0, 2-3) = (-1, 4, -1)\) \(\vec{AC} = C - A = (1-1, 1-0, 1-3) = (0, 1, -2)\)
02

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 \(\vec{n} = (1,1,1)\).
03

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 \(x+y+z=5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the ellipse \(x^{2}+4 y^{2}=1\) in the \(x y\) -plane. a. If this ellipse is revolved about the \(x\) -axis, what is the equation of the resulting ellipsoid? b. If this ellipse is revolved about the \(y\) -axis, what is the equation of the resulting ellipsoid?

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x-y)^{2}}{\left(x^{2}+y^{2}\right)^{3 / 2}}$$

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum.

Problems with two constraints Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0,\) and \(h(x, y, z)=0\)

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.
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