AC3 Chap 9 4 - Practice problem for chapter 9.4 PDF

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Practice problem for chapter 9.4...

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AC3 Chapter 9.4 Activities & Exercises Xinli Wang August 2020

Chapter 9.4 Activities 1.

• What are two unit vectors that are orthogonal to both i and j? • What is i × j? Explain. • Determine each of i × k

j × i

j × k

k × i

k × j

3

• Let v be a vector in R . What is v × v? 3

• Let u and v be parallel vectors in R . What is u × v? Explain. Solution: • Two unit vectors that are orthogonal to both i and j are k and −k. • The vectors i, j, and k form a right hand system, so i × j = k.

• The vectors i, k, and −j form a right hand system, so i × k = −j. Similarly, j × i = − k j × k = i    k ×i = j    k × j = −i. • Since the angle between v and v is 0, the definition tells us that 2 v × v = ∣ v∣ sin(0) n = 0.

• The angle between parallel vectors u and v is θ = 0 or θ = π, so the definition tells us that u × v = ∣u∣ ∣ v∣ sin(θ) n = 0.

2.

• Is the cross product commutative? That is, is u × v = v × u for any 3 vectors u and v in R ? If yes, explain why. If no, how are v × u and u × v related? Why? (Hint: Think about how u, v, and u × v are related.) 1

• Calculate (i × i) × j and i × (i × j).

• Is the cross product associative? That is, is u × ( v × w)  = ( u × v) × w 3 for any vectors u, v, and w in R ?

Solution: • The answer is no. If u, v, and n  form a right hand system, then v, u, and − n form a right hand system. So we have  so • We know that i × i = 0, (i × i) × j = 0 × j = 0.

However, i × j = k and so

i × ( j) =    i× i × k = −j. • The result of part (b) shows that the cross product is not associative. 3

3. Let u and v be vectors in R and let c be a scalar. Show that u × (c v) = c( u × v).

Solution: Anti-commutative property gives us that u × (c v) = −((c v) × u). Property of scalar multiplication gives us u × (c v) = −(c)( v × u). Applying anti-commutative property again shows that u × (c v) = −(c)(− u × v) = c( u × v).

4.

• Find the area of the parallelogram formed by the vectors u = ⟨1, 3, −2⟩ and v = ⟨3, 0, 1⟩. • Find the area of the parallelogram in R whose vertices are (1, 0, 1), (0, 0, 1), (2, 1, 0), and (1, 1, 0). 3

Solution:

• We just need to calculate ∣u × v∣:

√ ∣u × v∣ = ∣⟨3, −7, −9⟩∣ = 139.

• Let P = (1, 0, 1), Q = (0, 0, 1), R = (2, 1, 0), and S = (1, 1, 0). Two 3 vectors that determine this parallelogram in R are the vectors −−→ P Q = ⟨−1, 0, 0⟩ = −i 2

and

−−→ P R = ⟨1, 1, −1⟩ = i + j − k.

Now

−−→ −−→ P Q × P R = −i × (i + j − k) = −[(i × i) + (i × j) − (i × k)] = −[0 + k + j] = ⟨0, −1, −1⟩.

−−→ −−→ So the area of the parallelogram determined by P Q and P R is √ −−→ −−→ ∣P Q × P R∣ = ∣⟨0, −1, −1⟩∣ = 2.

5. Calculate

• ⟨2, −1, 0⟩ × ⟨0, 1, 3⟩.

• ⟨2, −1, 0⟩ × ⟨−4, 2, 0⟩. Why should you have expected this result?

• The volume of the parallelepiped determined by the vectors ⟨2, −1, 0⟩, ⟨0, 1, 3⟩, and ⟨1, 1, 1⟩.

Solution:

• We use formula (9.4.1) with ⟨u1 , u2 , u3 ⟩ = ⟨2, −1, 0⟩ and ⟨v1 , v2 , v3 ⟩ = ⟨0, 1, 3⟩ to obtain

⟨2, −1, 0⟩ × ⟨0, 1, 3⟩ = ((−1)(3) − (0)(1))i − ((2)(3) − (0)(0))j + ((2)(1) − (−1)(0))k = −3 i − 6j + 2 k.

• Similar to part (a) we have ⟨2, −1, 0⟩ × ⟨−4, 2, 0⟩ = (0)i − (0)j + (0) k = 0.

Since (−2)⟨2, −1, 0⟩ = ⟨−4, 2, 0⟩, these two vectors are parallel. So we should have expected to obtain a cross product of 0.

• The vectors u = ⟨2, −1, 0⟩, v = ⟨1, 1, 1⟩, and w = ⟨0, 1, 3⟩, and form a right hand system, so the volume of the parallelepiped determined by u, v, and w is  ⋅ (j + 3k) ( u × v) ⋅ w = (2 i − 2 j + 3k) = 7.

6. Suppose u = ⟨3, 5, −1⟩ and v = ⟨2, −2, 1⟩. • Find two unit vectors orthogonal to both u and v. 3

• Find the volume of the parallelepiped formed by the vectors u, v, and w = ⟨3, 3, 1⟩.

• Find a vector orthogonal to the plane containing the points (0, 1, 2), (4, 1, 0), and (−2, 2, 2). • Given the vectors u and v shown below in Figure 1, sketch the cross products u × v and v × u.

Figure 1: Vectors u and v • Do the vectors a  = ⟨1, 3, −2⟩, b = ⟨2, 1, −4⟩, and c = ⟨0, 1, 0⟩ lie in the same plane? Use the concepts from this section to explain. Solution: • The vector u × v is orthogonal to u and v. Now

u × v = ⟨3, 5, −1⟩ × ⟨2, −2, 1⟩ = ⟨3, −5, −16⟩,

so one unit vector orthogonal to u and v is z =

1 1 u × v = √ ⟨3, −5, −16⟩. ∣u × v∣ 290

Another unit vector orthogonal to u and v is − z.

• The volume of the parallelepiped formed by the u, v, and w is ∣( u × v) ⋅ w∣  = ∣⟨3, −5, −16⟩ ⋅ ⟨3, 3, 1⟩∣ = 22.

• Let P = (0, 1, 2), Q = (4, 1, 0), and R = (−2, 2, 2). Two vectors in −−→ −−→ the plane are P Q = ⟨4, 0, −2⟩ and P R = ⟨−2, 1, 0⟩. So the vector −−→ −−→ P Q × P R = ⟨2, 0, 4⟩ is a vector orthogonal to the plane containing the points. • Recall that u × v is orthogonal to u and v and that u, v and u × v in that order form a right hand system. • Note that u × v = ⟨−10, 0, −5⟩, and that u × v is orthogonal to every vector in the plane determined by u and v. Now ( u × v) ⋅ w = 0, so vecw is in the plane determined by u and v. 4

Chapter 9.4 Exercises 1. Let u = 2i + j and v = i + 2j be vectors in R3 . • Without doing any computations, find a unit vector that is orthogonal to both u and v. What does this tell you about the formula for u × v? • Using the bilinearity of the cross product and what you know about cross products involving the fundamental vectors i and j, compute u × v. • Next, use the determinant version of cross product to compute u × v. Write one sentence that compares your results in (a), (b), and (c). • Find the area of the parallelogram determined by u and v. Solution: • Since u and v lie in the x-y plane, the vector k is orthogonal to both. So u × v is a multiple of k. • The properties of the cross product tell us that u × v = (2 i + j) × (i + 2j) = 2(i × i) + (j × i) + 4(i × j) + 2(j × j) = 2(i × i) − (i × j) + 4(i × j) + 2(j × j) = 2(i × i) + 3(i × j) + 2(j × j) = 20 + 3k+ 20 = 3k. • Using the determinant version of u × v we find that »» »» »» »» »» »»

 i 2 1

 j 1 2

 k 0 0

»» »» »» = [(2)(0)−(0)(1)]i−[(1)(0)−(0)(2)]j+[(2)(2)−(1)(1)]k = 3 k. »» »» »»

The vectors u and v lie in the x-y plane and a unit vector that is orthogonal to both u and v is 3 k, established though the determinant formula for the cross product and through the properties of the cross product. • The area of the parallelogram determined by u and v is ∣u × v∣ = ∣3 k∣ = 3.

2. Let x = ⟨1, 1, 1⟩ and y = ⟨0, 3, −2⟩.

• Are x and y orthogonal? Are x and y parallel? Clearly explain how you know, using appropriate vector products. 5

• Find a unit vector that is orthogonal to both x and y. • Express y as the sum of two vectors: one parallel to x , the other orthogonal to x . • Determine the area of the parallelogram formed by x and y. Solution: • Two vectors are orthogonal if their dot product is 0. Since x ⋅ y = 1, the vectors x and y are not orthogonal. Vectors x and y are parallel if one is a scalar multiple of the other, or if their cross product is 0. Since x × y = −5i + 2 j + 3k, the vectors x and y are not parallel. • The vector x × y = −5i+2 j+3k is orthogonal to both x and y, so a unit 1 x × y = vector that is orthogonal to both x and y is ∣x×  y∣ 

• The projection

1 √ ⟨−5, 2, 3⟩. 38

y ⋅ x 1 x = √ ⟨1, 1, 1⟩ x ⋅ x 3 of y onto x is a vector that is parallel to x , while the vector projx y =

√ √ 1 y − proj x y = √ ⟨−1, 3 3 − 1, −2 3 − 1⟩ 3

is orthogonal to x . Note that y = proj x y + ( y − proj x y).  • The area of the parallelogram determined by x and y is √ ∣x × y∣  = ∣ − 5i + 2 j + 3 k∣ = 38.

3. Consider the triangle in R formed by P (3, 2, −1), Q(1, −2, 4), and R(4, 4, 0). 3

−−→ −−→ • Find P Q and P R.

• Observe that the area of △P QR is half of the area of the parallelo−−→ −−→ gram formed by P Q and P R. Hence find the area of △P QR. • Find a unit vector that is orthogonal to the plane that contains points P , Q, and R. • Determine the measure of ∠P QR. Solution: • With the given points P , Q, and R we have −−→ −−→ P Q = ⟨−2, −4, 5⟩ and P R = ⟨1, 2, 1⟩. −−→ −−→ • The area of the parallelogram formed by P Q and P R is √ −−→ −−→ ∣ P Q × P R∣ = ∣⟨−14, 7, 0⟩∣ = 245. Hence the area of △P QR is

√

6

245 . 2

−−→ −−→ • The cross product P Q × P R is orthogonal to the plane containing P , Q, and R, so a unit vector that is orthogonal to the plane that contains points P , Q, and R is −−→ −−→ 1 1 ⟨−14, 7, 0⟩. −−→ −−→ P Q × P R = √ 245 ∣ P Q × P R∣ • If θ is the measure of ∠P QR, then −−→ −−→ QP ⋅ QR ( −−→ −−→ ) ∣ QP ∣∣QR∣ ⟨2, 4, −5⟩ ⋅ ⟨3, 6, −4⟩ −1 = cos ( ) ∣⟨2, 4, −5⟩∣∣⟨3, 6, −4⟩∣ 50 −1 = cos ( √ √ ) 45 61 ◦ ≈ 17.38 . −1

θ = cos

Acknowledgement: These questions and solutions are part of Active Calculus– Multivariable by Steve Schlicker.

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