## Engage NY Eureka Math Precalculus Module 2 Lesson 26 Answer Key

### Eureka Math Precalculus Module 2 Lesson 26 Exercise Answer Key

Exercises

Exercise 1.

Describe the set of points (8t,3t), where t represents a real number.

Answer:

The point (8t,3t) is a dilation of the point (8,3) with a scale factor of t. Thus, each such point lies on the line that joins (8,3) to the origin.

Exercise 2.

Project the point (8,3) onto the line x = 1.

Answer:

We need the value of t that makes 8t = 1, so t = \(\frac{1}{8}\). Thus, 3t = 3(\(\frac{1}{8}\)) = \(\frac{3}{8}\), and so the image is (1,\(\frac{3}{8}\)).

Exercise 3.

Project the point (8,3) onto the line x = 5.

Answer:

We need the value of t that makes 8t = 5, so t = \(\frac{5}{8}\). Thus, 3t = 3(\(\frac{3}{8}\)) = \(\frac{15}{8}\), and so the image is (5,\(\frac{15}{8}\)).

Exercise 4.

Project the point (-1,4,5) onto the plane y = 1.

Answer:

We need the value of t that makes 4t = 1, so t = \(\frac{1}{4}\). Thus, the image is (-\(\frac{1}{4}\),1,\(\frac{5}{4}\)).

Exercise 5.

Project the point (9,5,-8) onto the plane z = 3.

Answer:

We need the value of t that makes -8t = 3, so t = –\(\frac{3}{8}\). Thus, the image is (-\(\frac{27}{8}\),-\(\frac{15}{8}\),3).

### Eureka Math Precalculus Module 2 Lesson 26 Problem Set Answer Key

Question 1.

A cube in 3-D space has vertices \(\left(\begin{array}{l}

10 \\

10 \\

10

\end{array}\right),\left(\begin{array}{l}

13 \\

10 \\

10

\end{array}\right),\left(\begin{array}{l}

10 \\

13 \\

10

\end{array}\right),\left(\begin{array}{l}

10 \\

10 \\

13

\end{array}\right),\left(\begin{array}{l}

13 \\

13 \\

10

\end{array}\right),\left(\begin{array}{l}

13 \\

10 \\

13

\end{array}\right),\left(\begin{array}{l}

10 \\

13 \\

13

\end{array}\right),\left(\begin{array}{l}

13 \\

13 \\

13

\end{array}\right)\)

a. How do we know that these vertices trace a cube?

Answer:

The edges between the vertices form right angles, and all have a length of 3 units.

b. What is the volume of the cube?

Answer:

3^{3} = 27 cubic units

c. Let z = 1. Find the eight points on the screen that represent the vertices of this cube (some may be obscured).

Answer:

\(\left(\begin{array}{l}

1 \\

1 \\

1

\end{array}\right),\left(\begin{array}{c}

1.3 \\

1 \\

1

\end{array}\right),\left(\begin{array}{c}

1 \\

1.3 \\

1

\end{array}\right),\left(\begin{array}{c}

\frac{1}{1.3} \\

\frac{1}{1.3} \\

1

\end{array}\right),\left(\begin{array}{c}

1.3 \\

1.3 \\

1

\end{array}\right),\left(\frac{1}{1.3}\right),\left(\begin{array}{c}

\frac{1}{1.3} \\

1 \\

1

\end{array}\right),\left(\begin{array}{l}

1 \\

1 \\

1

\end{array}\right)\)

d. What do you notice about your result in part (c)?

Answer:

Some of the points went to the same position. These were points that lay along the same path from the camera. Some points were dilated by a factor of \(\frac{1}{10}\) and some by a factor of \(\frac{1}{13}\) depending on how far they were from the plane in the z-direction.

Question 2.

An object in 3-D space has vertices \(\left(\begin{array}{l}

1 \\

5 \\

0

\end{array}\right),\left(\begin{array}{l}

0 \\

6 \\

0

\end{array}\right),\left(\begin{array}{l}

0 \\

5 \\

1

\end{array}\right),\left(\begin{array}{c}

-1 \\

5 \\

0

\end{array}\right),\left(\begin{array}{l}

0 \\

4 \\

0

\end{array}\right)\).

a. What kind of shape is formed by these vertices?

Answer:

It appears this is a pyramid with a square base.

b. Let y = 1. Find the five points on the screen that represent the vertices of this shape.

Answer:

\(\left(\begin{array}{c}

0.2 \\

1 \\

0

\end{array}\right),\left(\begin{array}{l}

0 \\

1 \\

0

\end{array}\right),\left(\begin{array}{c}

0 \\

1 \\

0.2

\end{array}\right),\left(\begin{array}{c}

-0.2 \\

1 \\

0

\end{array}\right),\left(\begin{array}{l}

0 \\

1 \\

0

\end{array}\right)\)

Question 3.

Consider the shape formed by the vertices given in Problem 2.

a. Write a transformation matrix that rotates each point around the y-axis Î¸ degrees.

Answer:

\(\left(\begin{array}{ccc}

\cos (\theta) & 0 & -\sin (\theta) \\

0 & 1 & 0 \\

\sin (\theta) & 0 & \cos (\theta)

\end{array}\right)\)

b. Project each rotated point onto the plane y = 1 if Î¸ = 45Â°.

Answer:

First we rotate each point: \(\left(\begin{array}{c}

\frac{\sqrt{2}}{2} \\

5 \\

\frac{\sqrt{2}}{2}

\end{array}\right),\left(\begin{array}{c}

0 \\

6 \\

0

\end{array}\right),\left(\begin{array}{c}

-\frac{\sqrt{2}}{2} \\

5 \\

\frac{\sqrt{2}}{2}

\end{array}\right),\left(\begin{array}{c}

-\frac{\sqrt{2}}{2} \\

5 \\

-\frac{\sqrt{2}}{2}

\end{array}\right),\left(\begin{array}{l}

0 \\

4 \\

0

\end{array}\right) .\).

c. Is this the same as rotating the values you obtained in Problem 2 by 45Â°?

Answer:

Yes. Since the projection is interpreted as a dilation, order does not matter.

Question 4.

In technical drawings, it is frequently important to preserve the scale of the objects being represented. To accomplish this, we use an orthographic projection instead of a perspective projection. The idea behind the orthographic projection is that the points are translated at right angles to the screen (the word stem ortho- means straight or right). To project onto the xy-plane for instance, we can use the matrix \(\left(\begin{array}{lll}

1 & 0 & 0 \\

0 & 1 & 0

\end{array}\right)\).

a. Project the cube in Problem 1 onto the xy-plane by finding the 8 points that correspond to the vertices.

Answer:

\(\left(\begin{array}{l}

10 \\

10

\end{array}\right),\left(\begin{array}{l}

13 \\

10

\end{array}\right),\left(\begin{array}{l}

10 \\

13

\end{array}\right),\left(\begin{array}{l}

10 \\

10

\end{array}\right),\left(\begin{array}{l}

13 \\

13

\end{array}\right),\left(\begin{array}{l}

13 \\

10

\end{array}\right),\left(\begin{array}{l}

10 \\

13

\end{array}\right),\left(\begin{array}{l}

13 \\

13

\end{array}\right)\)

b. What do you notice about the vertices of the cube after projecting?

Answer:

They form a square with all 8 points being mapped to 4 points.

c. What shape is visible on the screen?

Answer:

It is a square with sides of length 3.

d. Is the area of the shape that is visible on the screen what you expected from the original cube? Explain.

Answer:

The area of the visible shape is 9, which is the same as the area of any of the faces of the original cube.

e. Summarize your findings from parts (a)â€“(d).

Answer:

Orthographic projections preserve the scale of objects but much of the original information is lost.

f. State the orthographic projection matrices for the xz-plane and the yz-plane.

Answer:

For xz-plane: \(\left(\begin{array}{lll}

1 & 0 & 0 \\

0 & 0 & 1

\end{array}\right)\)

For yz-plane: \(\left(\begin{array}{lll}

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right)\)

g. In regard to the dimensions of the orthographic projection matrices, what causes the outputs to be two-dimensional?

Answer:

The transformation matrices are 2 Ã— 3, meaning they can operate on 3-dimensional points but generate 2-dimensional outputs.

Question 5.

Consider the point A = \(\left(\begin{array}{l}

a_{x} \\

a_{y} \\

a_{z}

\end{array}\right)\) in the field of view from the origin through the plane z = 1.

a. Find the projection of A onto the plane z = 1.

Answer:

Since the point is in the field of view, we know that a_{z}â‰ 0, and we get: \(\left(\begin{array}{l}

\frac{a_{x}}{a_{z}} \\

\frac{a_{y}}{a_{z}} \\

1

\end{array}\right)\).

b. Find a 3 Ã— 3 matrix P such that PA finds the projection of A onto the plane z = 1.

Answer:

\(\left(\begin{array}{ccc}

\frac{1}{a_{z}} & 0 & 0 \\

0 & \frac{1}{a_{z}} & 0 \\

0 & 0 & \frac{1}{a_{z}}

\end{array}\right)\)

c. How does the matrix change if instead of projecting onto z = 1, we project onto z = c, for some real number câ‰ 0?

Answer:

The numerator of each fraction is c instead of 1.

d. Find the scalars that generate the image of A onto the planes x = c and y = c, assuming the image exists. Describe the scalars in words.

Answer:

\(\frac{c}{a_{x}}\) and \(\frac{c}{a_{y}}\). Whenever we are projecting onto a plane parallel to the xy-, xz-, or yz-planes, the scalar by which we multiply the point A is always going to be the reciprocal of z-, y-, or x-coordinate times the z-, y-, or x-location of the plane.

Extension:

Question 6.

Instead of considering the rotation of a point about an axis, consider the rotation of the camera. Rotations of the camera cause the screen to rotate along with it, so that to the viewer, the screen appears immobile.

a. If the camera rotates Î¸_{x} around the x-axis, how does the computer world appear to move?

Answer:

A rotation of -Î¸_{x} about the x-axis

b. State the rotation matrix we could use on a point A to simulate rotating the camera and computer screen by Î¸_{x} about the x-axis but in fact keeping the camera and screen fixed.

Answer:

\(\left(\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left(-\theta_{x}\right) & -\sin \left(-\theta_{x}\right) \\

0 & \sin \left(-\theta_{x}\right) & \cos \left(-\theta_{x}\right)

\end{array}\right)\)

c. If the camera rotates Î¸_{y} around the y-axis, how does the computer world appear to move?

Answer:

A rotation of -Î¸_{y} about the y-axis

d. State the rotation matrix we could use on a point A to simulate rotating the camera and computer screen by Î¸_{y} about the y-axis but in fact keeping the camera and screen fixed.

Answer:

\(\left(\begin{array}{ccc}

\cos \left(-\theta_{y}\right) & 0 & -\sin \left(-\theta_{y}\right) \\

0 & 1 & 0 \\

\sin \left(-\theta_{y}\right) & 0 & \cos \left(-\theta_{y}\right)

\end{array}\right)\)

e. If the camera rotates Î¸_{z} around the z-axis, how does the computer world appear to move?

Answer:

A rotation of -Î¸_{z} about the z-axis

f. State the rotation matrix we could use on a point A to simulate rotating the camera and computer screen by Î¸_{z} about the z-axis but in fact keeping the camera and screen fixed.

Answer:

\(\left(\begin{array}{ccc}

\cos \left(-\theta_{z}\right) & -\sin \left(-\theta_{z}\right) & 0 \\

\sin \left(-\theta_{z}\right) & \cos \left(-\theta_{z}\right) & 0 \\

0 & 0 & 1

\end{array}\right)\)

g. What matrix multiplication could represent the camera starting at a relative angle (Î¸_x,Î¸_y,Î¸_z )? Apply the transformations in the order z-y-x. Do not find the product.

Answer:

\(\left(\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left(-\theta_{x}\right) & -\sin \left(-\theta_{x}\right) \\

0 & \sin \left(-\theta_{x}\right) & \cos \left(-\theta_{x}\right)

\end{array}\right)\left(\begin{array}{cccc}

\cos \left(-\theta_{y}\right) & 0 & -\sin \left(-\theta_{y}\right) \\

0 & 1 & 0 \\

\sin \left(-\theta_{y}\right) & 0 & \cos \left(-\theta_{y}\right)

\end{array}\right)\left(\begin{array}{ccc}

\cos \left(-\theta_{z}\right) & -\sin \left(-\theta_{z}\right) & 0 \\

\sin \left(-\theta_{z}\right) & \cos \left(-\theta_{z}\right) & 0 \\

0 & 0 & 1

\end{array}\right)\)

### Eureka Math Precalculus Module 2 Lesson 26 Exit Ticket Answer Key

Question 1.

Consider the plane defined by z = 2 and the points, x = \(\left(\begin{array}{l}

3 \\

6 \\

8

\end{array}\right)\) and y = \(\left(\begin{array}{c}

2 \\

-3 \\

5

\end{array}\right)\).

a. Find the projections of x and y onto the plane z = 2 if the eye is placed at the origin.

Answer:

x = \(\left(\begin{array}{l}

\frac{3}{4} \\

\frac{3}{2} \\

2

\end{array}\right)\), y = \(\left(\begin{array}{c}

\frac{4}{5} \\

-\frac{6}{5} \\

2

\end{array}\right)\)

b. Consider w = \(\left(\begin{array}{c}

\mathbf{1} \\

2 \\

-\mathbf{1}

\end{array}\right)\); does it make sense to find the projection of w onto z = 2? Explain.

Answer:

No, t = –\(\frac{1}{2}\).

Question 2.

Consider an object located at \(\left(\begin{array}{l}

3 \\

4 \\

0

\end{array}\right)\) and rotating around the z-axis. At what Î¸ value is the object out of sight of the plane y = 1?

Answer:

To rotate around the z-axis and project onto y = 1, \(\left(\begin{array}{ccc}

\cos (\theta) & -\sin (\theta) & 0 \\

\sin (\theta) & \cos (\theta) & 0 \\

0 & 0 & 1

\end{array}\right)\left(\begin{array}{l}

3 \\

4 \\

0

\end{array}\right) = \left(\begin{array}{c}

3 \cos (\theta)-4 \sin (\theta) \\

3 \sin (\theta)+4 \cos (\theta) \\

0

\end{array}\right)\)

To project onto y = 1, t = \(\frac{1}{3 \sin (\theta)+4 \cos (\theta)}\). The projection would be \(\left(\begin{array}{c}

\frac{3 \cos (\theta)-4 \sin (\theta)}{3 \sin (\theta)+4 \cos (\theta)} \\

1 \\

0

\end{array}\right)\). The value of Î¸ that would make the object out of site is when \(\frac{3 \cos (\theta)-4 \sin (\theta)}{3 \sin (\theta)+4 \cos (\theta)}\) = 0.

This occurs when 3â€†cos(Î¸)-4â€†sin(Î¸) = 0 or when Î¸ = \(\tan ^{-1}\left(\frac{3}{4}\right)\).