## Engage NY Eureka Math Precalculus Module 2 End of Module Assessment Answer Key

### Eureka Math Precalculus Module 2 End of Module Assessment Answer Key

Question 1.

a. Find values for a, b, c, d, and e so that the following matrix product equals the 3 × 3 identity matrix. Explain how you obtained these values.

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

a & -3 & 5 \\

c & c & 1 \\

5 & b & -4

\end{array}\right]\left[\begin{array}{ccc}

1 & b & d \\

1 & c & e \\

2 & b & b

\end{array}\right]\)

Answer:

The row 1, column 1 entry of the product is a – 3 + 10. This should equal 1, so a = -6.

The row 2, column 1 entry of the product is c + c + 2. This should equal zero, so c = -1.

The row 3, column 1 entry of the product is 5 + b – 8. This should equal zero, so b = 3.

The row 1, column 3 entry of the product is ad – 3e + 5b = -6d – 3e + 15, and this should equal zero. So, we should have 6d + 3e = 15.

The row 2, column 3 entry of the product is cd + ce + b = -d – e + 3, and this should equal zero. So, we need d + e = 3.

Solving the two linear equations d and e gives d = 2 and e = 1.

So, in summary, we have a = -6, b = 3, c = -1, d = 2, and e = 1.

b. Represent the following system of linear equations as a single matrix equation of the form Ax = b, where A is a 3 × 3 matrix, and x and b are 3 × 1 column matrices.

x + 3y + 2z = 8

x – y + z = -2

2x + 3y + 3z = 7

Answer:

We have Ax = b with A = \(\left[\begin{array}{ccc}

1 & 3 & 2 \\

1 & -1 & 1 \\

2 & 3 & 3

\end{array}\right]\) and x = \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\) and b = \(\left[\begin{array}{c}

8 \\

-2 \\

7

\end{array}\right]\).

c. Solve the system of three linear equations given in part (b).

Answer:

The solution is given by x = A^{-1}b, if the matrix inverse exists. But part (a) shows that if we set

B = \(\left[\begin{array}{ccc}

-6 & -3 & 5 \\

-1 & -1 & 1 \\

5 & 3 & -4

\end{array}\right]\),

then we have BA = 1. We could check that AB = 1, as well (in which case B is the matrix inverse of A), but even without knowing this, from

Question 2.

The following diagram shows two two-dimensional vectors v and w in the place positioned to each have an endpoint at point P.

a. On the diagram, make reasonably accurate sketches of the following vectors, again each with an endpoint at P. Be sure to label your vectors on the diagram.

i. 2v

ii. -w

iii. v + 3w

iv. w – 2v

v. \(\frac{1}{2}\)v

Answer:

Vector v has magnitude 5 units, w has magnitude 3, and the acute angle between them is 45°.

b. What is the magnitude of the scalar multiple -5v?

Answer:

We have ||-5v|| = 5||v|| = 5 × 5 = 25

c. What is the measure of the smallest angle between -5v and 3w if these two vectors are placed to have a common endpoint?

Answer:

The two vectors in question have an angle of smallest measure 135° between them.

Question 3.

Consider the two-dimensional vectors v = (2, 3) and w = (-2, -1).

a. What are the components of each of the vectors y + w and y – w?

Answer:

v + w = (2 + (-2) , 3 + (-1))= (0, 2) and v – w = (2- (-2) , 3 – (-1)) = (4, 4)

b. On the following diagram, draw representatives of each of the vectors v, w, and v + w, each with an endpoint at the origin.

Answer:

c. The representatives for the vectors v and w you drew form two sides of a parallelogram, with the vector v + w corresponding to one diagonal of the parallelogram. What vector, directed from the third quadrant to the first quadrant, is represented by the other diagonal of the parallelogram? Express your answer solely in terms of v and w, and also give the coordinates of this vector.

Answer:

Label the points A and B as shown. The vector we seek is \(\overrightarrow{A B} .\). To move from A to B we need to follow -w and then v. Thus, the vector we seek is -w + v, which is the same as v – w.

Also, we see that to move from A to B we need to move 4 units to the right and 4 units upward. This is consistent with v – w = (2 – (-2) , 3 – (-1)) = (4, 4).

d. Show that the magnitude of the vector v + w does not equal the sum of the magnitudes of v and of w.

Answer:

We have ||v|| = \(\sqrt{2^{2}+3^{2}}=\sqrt{13}\) and ||w|| = \(\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{5}\), so ||v|| + ||w|| = √13 + √5. Now, ||v + w|| = \(\sqrt{0^{2}+2^{2}}\) = 2. This does not equal √13 + √5.

e. Give an example of a non-zero vector u such that ||v + u|| does equal ||v|| + ||u||.

Answer:

Choosing u to be the vector v works.

||v + u|| = ||v + v|| = ||2v|| = 2||v|| = ||v|| + ||v||

(In fact, u = kv for any positive real number k words.)

(f Fact, u kv For aij positive real r.uivber k works.)

f. Which of the following three vectors has the greatest magnitude: v + (-w), w – v, or (-v) – (-w)?

Answer:

(-v) – (-w) = -v + w = w – v and

v + (-w) = v – w = -(w – v).

So, each of these vectors is either w – v or the scalar multiple (-1)(w – v), which is the same vector but with opposite direction. They all have the same magnitude.

g. Give the components of a vector one-quarter the magnitude of vector v and with direction opposite the direction of v.

Answer:

\(-\frac{1}{4}\)V = –\(\frac{1}{4}[/latex(2, 3) = (-[latex]\frac{1}{2}\), –\(\frac{3}{4}\))

Question 4.

Vector a points true north and has magnitude 7 units. Vector b points 30° east of true north. What should the magnitude of b be so that b – a points directly east?

a. State the magnitude and direction of b – a.

Answer:

We hope to have the following vector diagram incorporating a right triangle.

we have ||a|| = 7, and for this 30 – 60 – 90 triangle we need ||b|| = 2||b – a|| and 7 = ||a|| = √3||b – a||. This shows the magnitude b should be ||b|| = \(\frac{14}{\sqrt{3}}\).

The vector b – a points east and has magnitude \(\frac{7}{\sqrt{3}}\)

b. Write b – a magnitude and direction form.

Answer:

(\(\frac{7}{\sqrt{3}}\), 0°); b – a points east and has magnitude of \(\frac{7}{\sqrt{3}}\) and a direction of 0° measured from the horizontal.

Question 5.

Consider the three points A = (10, -3, 5), B = (0, 2, 4), and C = (2, 1, 0) in three-dimensional space. Let M be the midpoint of \(\overline{A B}\) and N be the midpoint of \(\overline{A C}\).

a. Write down the components of the three vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\), and \(\overrightarrow{C A}\), and verify through arithmetic that their sum is zero. Also, explain why geometrically we expect this to be the case.

Answer:

We have the following:

\(\overrightarrow{A B}\) = (-10, 5, 1)

\(\overrightarrow{B C}\) = (2, -1, -4)

\(\overrightarrow{C A}\) = (8, -4, 5)

Their sum is (-10 + 2 + 8, 5 – 1 – 4, -1 – 4 + 5) = (0, 0, 0).

This is to be expected as the three points A, B, and C are vertices of a triangle (even in three-dimensional space), and the vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\), and \(\overrightarrow{C A}\), when added geometrically, traverse the sides of the triangle and have the sum effect of “returning to start.” That is, the cumulative effect of the three vectors is no vectorial shift at all.

b. Write down the components of the vector \(\overrightarrow{M N}\). Show that it is parallel to the vector \(\overrightarrow{B C}\) and half its magnitude.

Answer:

We have M = (5, –\(\frac{1}{2}\), –\(\frac{9}{2}\)) and N = (6, -1, \(\frac{5}{2}\)). Thus \(\overrightarrow{M N}\) = (1, –\(\frac{1}{2}\), -2).

We see that \(\overrightarrow{M N}\) = \(\frac{1}{2}\) \(\overrightarrow{B C}\), which shows that \(\overrightarrow{M N}\) has the same direction as \(\overrightarrow{B C}\) (and hence is parallel to it) and half the magnitude.

Let G = (4, 0, 3).

c. What is the value of the ratio \(\frac{\|\overrightarrow{M G}\|}{\|\overrightarrow{M C}\|}\)?

Answer:

\(\overrightarrow{M G}\) = (-1, \(\frac{1}{2}\), –\(\frac{3}{2}\)) and \(\overrightarrow{M C}\) = (-3, \(\frac{3}{2}\), –\(\frac{9}{2}\)) = 3\(\overrightarrow{M G}\). Thus \(\frac{\|\overrightarrow{M G}\|}{\|\overrightarrow{M C}\|}=\frac{\|\overline{M G}\|}{3\|\overrightarrow{M C}\|}=\frac{1}{3}\)

d. Show that the point G lies on the line connecting M and C. Show that G also lies on the line connecting N and B.

Answer:

That \(\overrightarrow{M G}\) = \(\frac{1}{3}\)\(\overrightarrow{M C}\) means that the point G lies a third of the way along \(\overrightarrow{M C}\).

Check:

\(\overrightarrow{N G}\) = (-2, 1, \(\frac{1}{2}\)) = \(\frac{1}{3}\)(-6, 3, \(\frac{3}{2}\)) = \(\frac{1}{3}\)\(\overrightarrow{N B}\)

So, G also lies on \(\overrightarrow{N B}\) (and one-third of the way along, too).

Question 6.

A section of a river, with parallel banks 95 ft. apart, runs true north with a current of 2 ft/sec. Lashana, an expert swimmer, wishes to swim from point A on the west bank to the point B directly opposite it. In still water, she swims at an average speed of 3 ft/sec.

The diagram to the right illustrates the situation. To counteract the current, Lashana realizes that she is to swim at some angle θ to the east/west direction as shown.

With the simplifying assumptions that Lashana’s swimming speed will be a constant 3 ft/sec and that the current of the water is a uniform 2 ft/sec flow northward throughout all regions of the river (e.g., we ignore the effects of drag at the river banks,), at what angle θ to east/west direction should Lashana swim in order to reach the opposite bank precisely at point B? How long will her swim take?

a. What is the shape of Lashana’s swimming path according to an observer standing on the bank watching her swim? Explain your answer in terms of vectors.

Answer:

Lashana’s velocity vector v has magnitude 3 and resolves into two components as shown, a component in the east direction v_{E} and a component in the south direction v_{S}.

We see

||v_{S}|| = ||v||sin(θ) = 3sin(θ).

Lashana needs this component of her velocity vector to counteract the northward current of the water. This will ensure that Lashana will swim directly toward point B with no sideways deviation.

Since the current is 2ft/sec, we need 3sinθ = 2, showing that θ = sin^{-1}(\(\frac{2}{3}\)) ≈ 41.8°.

Lashana will then swim at a speed of ||v_{E}|| = 3cosθ ft/sec towards the opposite bank.

Since sin(θ) = \(\frac{2}{3}\), θ is part of a 2 – √5 – 3 right triangle, so cos(θ) = \(\frac{\sqrt{5}}{3}\). Thus ||v_{E}|| = √5 ft/sec.

She needs to swim an east/west distance of 95 ft. at this speed. It will take her \(\frac{95 \mathrm{ft}}{\sqrt{5} \mathrm{ft} / \mathrm{sec}}\) = 19√5 seconds ≈ 42 seconds to do this.

b. If the current near the banks of the river is significantly less than 2 ft/sec, and Lashana swims at a constant speed of 3 ft/sec at the constant angle θ to the east/west direction as calculated in part (a), will Lashana reach a point different from B on the opposite bank? If so, will she land just north or just south of B? Explain your answer.

Answer:

As noted in the previous solution, Lashana will have no sideways motion in her swim. She will swim a straight-line path from A to B.

If the current is slower than 2 ft/sec at any region of the river surface, Lashana’s velocity vector component v_{s}, which has magnitude 2 ft/sec, will be larger in magnitude than the magnitude of the current. Thus, she will swim slightly southward during these periods. Consequently, she will land at a point on the opposite bank south of B

Question 7.

A 5 kg ball experiences a force due to gravity \(\vec{F}\) of magnitude 49 newtons directed vertically downward. If this ball is placed on a ramp tilted at an angle of 45°, what is the magnitude of the component of this force, in newtons, on the ball directed 45° toward the bottom of the ramp? (Assume the ball has a radius small enough that all forces are acting at the point of contact of the ball with the ramp.)

Answer:

The force vector can be resolved into two components as shown: F_{ramp} and F_{perp}.

We are interested in the component F_{ramp}.

We see a 45- 90 – 45 triangle in this diagram, with hypotenuse of magnitude 49 N. This means that the magnitude of F_{ramp} is \(\frac{49 \mathrm{~N}}{\sqrt{2}}\) ≈ 35 N.

Question 8.

Let A be the point (1, 1, -3) and B be the point (-2, 1, -1) in three-dimensional space.

A particle moves along the straight line through A and B at uniform speed in such a way that at time t = 0 seconds the particle is at A, and at t = 1 second the particle is at B. Let P(t) be the location of the particle at time t (so, P(0) = A and P(1) = B).

a. Find the coordinates of the point P(t) each in terms of t.

Answer:

We will write the coordinates of points as 3 × 1 column matrices, as is consistent for work with matrix notation.

The velocity vector of the particle is \(\overrightarrow{A B}\) = (-3, 0, 2). So, its position at time t is

P(t) = A + t\(\overrightarrow{A B}\) = \(\left[\begin{array}{l}

1 \\

1 \\

-3

\end{array}\right]\) + t\(\left[\begin{array}{c}

-3 \\

0 \\

2

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

1-3 t \\

1 \\

-3+2 t

\end{array}\right]\)

b. Give a geometric interpretation of the point P(0.5).

Answer:

Since P(0) = A and P(1) = B, P(0.5) is the midpoint of \(\overline{A B}\).

Let L be the linear transformation represented by the 3 × 3 matrix \(\left[\begin{array}{lll}

2 & 0 & 1 \\

1 & 3 & 0 \\

0 & 1 & 1

\end{array}\right]\), and let \(A^{\prime}\) = LA and \(B^{\prime}\) = LB be the images of the points A and B, respectively, under L.

c. Find the coordinates of \(A^{\prime}\) and \(B^{\prime}\).

Answer:

\(A^{\prime}\) = \(\left[\begin{array}{lll}

2 & 0 & 1 \\

1 & 3 & 0 \\

0 & 1 & 1

\end{array}\right]\left[\begin{array}{c}

1 \\

1 \\

-3

\end{array}\right]=\left[\begin{array}{c}

-1 \\

4 \\

-2

\end{array}\right]\) and

\(B^{\prime}\) = \(\left[\begin{array}{lll}

2 & 0 & 1 \\

1 & 3 & 0 \\

0 & 1 & 1

\end{array}\right]\left[\begin{array}{c}

-2 \\

1 \\

-1

\end{array}\right]=\left[\begin{array}{c}

-5 \\

1 \\

0

\end{array}\right]\)

A second particle moves through three-dimensional space. Its position at time t is given by L(P(t)), the image of the location of the first particle under the transformation L.

d. Where is the second particle at times t = 0 and t = 1? Briefly explain your reasoning.

Answer:

We see L(P(0)) = L(A) = \(A^{\prime}\) and L(P(1)) = L(B) = \(B^{\prime}\). Since the position of the particle at time t is given by L(P(t)), to find the location at t = 0 and t = 1, evaluate L(P(0)) and L(P(1)).

e. Prove that the second particle is also moving along a straight-line path at uniform speed.

Answer:

At time t the location of the second particle is

L(P(t)) = \(\left[\begin{array}{ccc}

2 & 0 & 1 \\

1 & 3 & 0 \\

0 & 1 & 1

\end{array}\right]\left[\begin{array}{c}

1-3 t \\

1 \\

-3+2 t

\end{array}\right]=\left[\begin{array}{c}

-1-4 t \\

4-3 t \\

-2+2 t

\end{array}\right]=\left[\begin{array}{c}

-1 \\

4 \\

-2

\end{array}\right]+t\left[\begin{array}{c}

-4 \\

-3 \\

2

\end{array}\right]\)

We recognize this as

L(P(t)) = \(A^{\prime}\)+ t \(\overline{\mathrm{A}^{\prime} B^{\prime}}\).

Thus, the second particle is moving along the straight line through \(A^{\prime}\) and \(A^{\prime}\) at a uniform velocity given by the vector \(\overline{\mathrm{A}^{\prime} B^{\prime}}\) = (-4, -3, 2).