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Chapter 3: Problem 3

Find a point that lies on the plane \(x+3 y+5 z=9\) and a normal vector to the plane.

Short Answer

Expert verified
Point: (9, 0, 0); Normal vector: 鉄�1, 3, 5鉄�.

Step by step solution

01

Understanding the Problem

We are given a plane equation: \(x + 3y + 5z = 9\). The task requires us to find both a point on this plane and the normal vector to this plane.
02

Identify the Normal Vector

The coefficients of \(x\), \(y\), and \(z\) in the plane equation represent the components of a normal vector. Therefore, the normal vector is \( \langle 1, 3, 5 \rangle \).
03

Finding a Point on the Plane

Choose values for \(y\) and \(z\) to solve for \(x\). Let \(y = 0\) and \(z = 0\). Substitute these into the equation: \(x + 3(0) + 5(0) = 9\). This simplifies to \(x = 9\). So, the point \((9, 0, 0)\) lies on the plane.

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Key Concepts

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

Normal Vector
In coordinate geometry, a normal vector is a crucial concept when working with planes. For any plane defined by an equation like \( ax + by + cz = d \), the normal vector can be extracted directly from the coefficients \( a \), \( b \), and \( c \). This means that the vector is perpendicular to the plane surface.
  • The normal vector provides important orientation information about the plane. In our case, with the plane equation \( x + 3y + 5z = 9 \), the normal vector is \( \langle 1, 3, 5 \rangle \).
  • Knowing the normal vector is fundamental in problems dealing with angles or distances between planes or other geometric figures.
  • In more advanced applications, such as computer graphics, the normal vector helps in determining how light interacts with surfaces.
Plane Equation
Understanding the equation of a plane is essential in multivariable calculus. A plane in a three-dimensional space is defined by an equation of the form \( ax + by + cz = d \). This equation establishes the relationship between the variables \( x \), \( y \), and \( z \), representing points on the plane.
  • The coefficients \( a \), \( b \), and \( c \) are important as they indicate the orientation of the plane through the normal vector \( \langle a, b, c \rangle \).
  • The constant \( d \) doesn't affect the direction of the plane, but it defines its position relative to the origin.
By manipulating this equation, you can find specific points that lie on the plane or understand the plane's relation to other geometric elements, like lines or other planes. For instance, to find a point on the plane \( x + 3y + 5z = 9 \), you can assign any two of the variables while solving for the third, as shown with the point \( (9, 0, 0) \).
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This field blends algebra with geometric concepts, allowing us to represent geometric shapes, like lines and planes, using algebraic equations.
  • It provides powerful tools to solve geometric problems more accurately and efficiently using equations and coordinates, instead of pure construction.
  • Coordinate geometry forms the foundation for further studies in calculus, physics, and engineering by describing spatial relationships and transformations.
In the context of planes, coordinate geometry allows for easy calculation of distances, angles, and intersections with other objects in three-dimensional space. This interaction between algebra and geometry simplifies complex geometrical transformations and aids in visualizing mathematical concepts.

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Most popular questions from this chapter

Let \(U\) be an open set in \(\mathbb{R}^{n}\), and let \(f: U \rightarrow \mathbb{R}\) be a function that is continuous at a point a of \(U\) (a) If \(f(\mathbf{a})>0\), show that there exists an open ball \(B=B(\mathbf{a}, r)\) centered at a such that \(f(\mathbf{x})>\frac{f(\mathbf{a})}{2}\) for all \(\mathbf{x}\) in \(B .\left(\right.\) Hint : Let \(\left.\epsilon=\frac{f(\mathbf{a})}{2} .\right)\) (b) Similarly, if \(f(\mathbf{a})<0\), show that there exists an open ball \(B=B(\mathbf{a}, r)\) such that \(f(\mathbf{x})<\frac{f(\mathbf{a})}{2}\) for all \(\mathbf{x}\) in \(B\). In particular, if \(f(\mathbf{a}) \neq 0\), there is an open ball \(B\) centered at a throughout which \(f(\mathbf{x})\) has the same sign as \(f(\mathbf{a})\).

The graph of the function \(f(x, y)=x^{3}-3 x y^{2}\) is called a monkey saddle. You will need to bring one along whenever you invite a monkey to go riding with you. The level sets of \(f\) are not easy to sketch directly, but it is still possible to get a reasonable idea of what the graph looks like. (a) Sketch the level set corresponding to \(c=0 .\) (Hint: \(\left.x^{3}-3 x y^{2}=x(x+\sqrt{3} y)(x-\sqrt{3} y) .\right)\) (b) Draw the region of the \(x y\) -plane in which \(f(x, y)>0\) and the region in which \(f(x, y)<0\). (Hint: Part (a) might help.) (c) Use the information from parts (a) and (b) to make a rough sketch of the monkey saddle.

Let \(\mathbf{v}\) and \(\mathbf{w}\) be vectors in \(\mathbb{R}^{n}\). (a) Show that \(\|\mathbf{v}\|-\|\mathbf{w}\| \leq\|\mathbf{v}-\mathbf{w}\| .(\) Hint: \(\mathbf{v}=(\mathbf{v}-\mathbf{w})+\mathbf{w} .)\) (b) Show that \(|\|\mathbf{v}\|-\|\mathbf{w}\|| \leq\|\mathbf{v}-\mathbf{w}\|\).

Sketch the surface in \(\mathbb{R}^{3}\) described by the given equation. $$ x^{2}+z^{2}=4 $$

Let \(U\) be the set of points \((x, y)\) in \(\mathbb{R}^{2}\) that satisfy the given conditions. Sketch \(U,\) and determine whether it is an open set. Your arguments should be at a level of rigor comparable to those given in the text. $$ 1
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