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Chapter 12: Problem 58

Describe with a sketch the sets of points \((x, y, z)\) satisfying the following equations. $$(x+1)(y-3)=0$$

Short Answer

Expert verified
Question: What is the intersection of the sets of points satisfying the equation \((x+1)(y-3)=0\), and describe its orientation and position in the 3D coordinate space. Answer: The intersection of the sets of points satisfying the equation \((x+1)(y-3)=0\) is a line parallel to the z-axis and passes through the point (-1, 3, z), where z is any value along the z-axis.

Step by step solution

01

Rewrite the equation

To better understand the geometry involved, let's rewrite the given equation: $$(x+1)(y-3)=0$$ This equation is a product of two linear expressions, and it equals 0 if either of the expressions equals 0. Therefore, we have two equations: $$x+1 = 0 \text{ or } y-3 = 0$$ These represent two planes in the three-dimensional space, with the first equation representing a vertical plane parallel to the yz-plane and the second representing a horizontal plane parallel to the xz-plane.
02

Determine the plane equation

For the first equation, \(x+1=0\), we can write it as the standard equation for a plane: $$x - (-1) = 0$$ This plane has a normal vector of \((1,0,0)\) and passes through the point \((-1,0,0)\). For the second equation, \(y-3=0\), we can write it as the standard equation for a plane: $$y-3=0$$ This plane has a normal vector of \((0,1,0)\) and passes through the point \((0,3,0)\).
03

Sketch the two planes

Now that we know the equations, we can sketch the two planes defined by these equations. The first plane, defined by \(x+1=0\), is a vertical plane parallel to the yz-plane, passing through x = -1. The second plane, defined by \(y-3=0\), is a horizontal plane parallel to the xz-plane, passing through y = 3.
04

Visualize the intersection of the two planes

We have now sketched and identified the two planes. The sets of points satisfying the given equation correspond to the points where these two planes intersect. In a three-dimensional space, the intersection of two distinct planes forms a line. In this case, the intersection is a line parallel to the z-axis that passes through the point \((-1, 3, z)\). To summarize, the sets of points \((x, y, z)\) satisfying the equation \((x+1)(y-3)=0\) correspond to the points in the intersection of two planes, which forms a line parallel to the z-axis passing through the point (-1, 3, z).

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Most popular questions from this chapter

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned} \mathbf{r}(t)=&(2 \cos t+2 \sin t) \mathbf{i}+(-\cos t+2 \sin t) \mathbf{j} \\\ &+(\cos t-2 \sin t) \mathbf{k} \end{aligned}$$

Zero curvature Prove that the curve $$ \mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle $$ where \(a, b, c, d, e,\) and \(f\) are real numbers and \(p\) is a positive integer, has zero curvature. Give an explanation.

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-1,2,3\rangle, \mathbf{v}=\langle 2,1,1\rangle\)

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