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Chapter 12: Problem 16

Sketch the following planes in the window \([0,5] \times[0,5] \times[0,5]\) $$z=3$$

Short Answer

Expert verified
Question: Sketch the plane z=3 within a 3D window of x, y, and z ranging from 0 to 5 and describe its position relative to the xy plane. Answer: The plane z=3 is a horizontal plane parallel to the xy plane, located at a distance of 3 from the xy plane. It intersects the 3D window at points (5, 0, 3), (0, 5, 3), and (0, 0, 3).

Step by step solution

01

Set up a 3D coordinate system

Create a 3D coordinate system with x, y, and z axes. Label the axes accordingly and mark the axes with the appropriate range (\([0,5]\) in this case).
02

Identify the plane equation and boundaries

The given plane equation is \(z=3\). We need to sketch this plane within a window of \(x = [0, 5]\), \(y = [0, 5]\), and \(z = [0, 5]\).
03

Plot the vertices of the window and the intersection points

Plot the vertices of the window in the 3D coordinate system: \((0, 0, 0), (5, 0, 0), (0, 5, 0), (0, 0, 5),\) \((5, 5, 0), (5, 0, 5),\) and \((0, 5, 5)\) Using the equation \(z = 3\), we can find the intersection points with the window edges: \((5, 0, 3), (0, 5, 3), (0, 0, 3),\) \((5, 5, 3),\) \((5, 0, 3),\) and \((0, 5, 3)\)
04

Sketch the plane

Connect the intersection points \((5, 0, 3), (0, 5, 3),\) and \((0, 0, 3)\) to their corresponding vertices in the window. Then, sketch the plane \(z=3\) by connecting the intersection points \((5, 0, 3), (0, 5, 3),\) and \((0, 0, 3)\). The plane will be parallel to the xy plane with a distance of 3 from the xy plane.
05

Label the sketch

Label the intersection points and make sure the coordinate axes are properly labeled with their respective ranges. Also, include the equation of the plane, \(z=3\), on the sketch.

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