# 18.2: Vectors

- Page ID
- 7941

## Check Your Understanding

**2.1.** a. Not equal because they are orthogonal; b. not equal because they have different magnitudes; c. not equal because they have different magnitudes and directions; d. not equal because they are antiparallel; e. equal.

**2.2. **16 m; \(\vec{D}\) = −16 m \(\hat{u}\)

**2.3. **G = 28.2 cm, \(\theta_{G}\) = 291°

**2.4. **\(\vec{D}\) = (−5.0 \(\hat{i}\) − 3.0 \(\hat{j}\))cm; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.

**2.5. **5.83 cm, 211°

**2.6.** \(\vec{D}\) = (−20 m) \(\hat{j}\)

**2.7.** 35.1 m/s = 126.4 km/h

**2.8. **\(\vec{G}\) = (10.25 \(\hat{i}\) − 26.22 \(\hat{j}\))cm

**2.9. **D = 55.7 N; direction 65.7° north of east

**2.10. **\(\hat{v}\) = 0.8 \(\hat{i}\) + 0.6 \(\hat{j}\), 36.87° north of east

**2.11. **\(\vec{A} \cdotp \vec{B}\) = −57.3, \(\vec{F} \cdotp \vec{C}\) = 27.8

**2.13. **131.9°

**2.14.** W_{1} = 1.5 J, W_{2} = 0.3 J

**2.15.** \(\vec{A} \times \vec{B}\) = −40.1 \(\hat{k}\) or, equivalently, |\(\vec{A} \times \vec{B}\)| = 40.1, and the direction is into the page; \(\vec{C} \times \vec{F}\) = + 157.6 \(\hat{k}\) or, equivalently, |\(\vec{C} \times \vec{F}\)| = 157.6, and the direction is out of the page.

**2.16.** a. −2 \(\hat{k}\), b. 2, c. 153.4°, d. 135°

## Conceptual Questions

**1.** Scalar

**3.** Answers may vary

**5. **Parallel, sum of magnitudes, antiparallel, zero

**7.** Yes, yes

**9. **Zero, yes

**11. **No

**13. **Equal, equal, the same

**15. **A unit vector of the x-axis

**17. **They are equal.

**19. **Yes

**21.** a. C = \(\vec{A} \cdotp \vec{B}\), b. \(\vec{C} = \vec{A} \times \vec{B}\) or \(\vec{C} = \vec{A} - \vec{B}\), c. \(\vec{C} = \vec{A} \times\vec{B}\), d. \(\vec{C}\) = A\(\vec{B}\), e. \(\vec{C} + 2 \vec{A} = \vec{B}\), f. \(\vec{C} = \vec{A} \times \vec{B}\), g. left side is a scalar and right side is a vector, h. \(\vec{C} = 2 \vec{A} \times \vec{B}\), i. \(\vec{C} = \frac{\vec{A}}{B}\), j. \(\vec{C} = \frac{\vec{A}}{B}\)

**23.** They are orthogonal.

## Problems

**25. **\(\vec{h}\) = −16.4 m \(\hat{u}\), 16.4 m

**27.** 30.8 m, 35.7° west of north

**29.** 134 km, 80°

**31. **7.34 km, 63.5° south of east

**33. **3.8 km east, 3.2 km north, 7.0 km

**35. **14.3 km, 65°

**37. **a. \(\vec{A}\) = + 8.66 \(\hat{i}\) + 5.00 \(\hat{j}\)

b. \(\vec{B}\) = + 3.01 \(\hat{i}\) + 3.99 \(\hat{j}\)

c. \(\vec{C}\) = + 6.00 \(\hat{i}\) − 10.39 \(\hat{j}\)

d. \(\vec{D}\) = −15.97 \(\hat{i}\) + 12.04 \(\hat{j}\)

f. \(\vec{F}\) = −17.32 \(\hat{i}\) − 10.00 \(\hat{j}\)

**39.** a. 1.94 km, 7.24 km

b. proof

**41.** 3.8 km east, 3.2 km north, 2.0 km, \(\vec{D}\) = (3.8 \(\hat{i}\) + 3.2 \(\hat{j}\))km

**43. **P_{1}(2.165 m, 1.250 m), P_{2}(−1.900 m, 3.290 m), 5.27 m

**45. **8.60 m, A(2\(\sqrt{5}\) m, 0.647\(\pi\)), B(3\(\sqrt{2}\) m, 0.75\(\pi\))

**47. **a. \(\vec{A} + \vec{B}\) = −4 \(\hat{i}\) − 6 \(\hat{j}\), |\(\vec{A} + \vec{B}\)| = 7.211, \(\theta\) = 236.3°

b. \(\vec{A} -\vec{B}\) = -2 \(\hat{i}\) + 2 \(\hat{j}\), |\(\vec{A} - \vec{B}\)| = 2\(\sqrt{2}\), \(\theta\) = 135°

**49. **a. \(\vec{C}\) = (5.0 \(\hat{i}\) − 1.0 \(\hat{j}\) − 3.0 \(\hat{k}\))m, C = 5.92 m

b. \(\vec{D}\) = (4.0 \(\hat{i}\) − 11.0 \(\hat{j}\) + 15.0 \(\hat{k}\))m, D = 19.03 m

**51.** \(\vec{D}\) = (3.3 \(\hat{i}\) − 6.6 \(\hat{j}\))km, \(\hat{i}\) is to the east, 7.34 km, −63.5°

**53. **a. \(\vec{R}\) = −1.35 \(\hat{i}\) − 22.04 \(\hat{j}\)

b. \(\vec{R}\) = −17.98 \(\hat{i}\) + 0.89 \(\hat{j}\)

**55. **\(\vec{D}\) = (200 \(\hat{i}\) + 300 \(\hat{j}\))yd, D = 360.5 yd, 56.3° north of east; The numerical answers would stay the same but the physical unit would be meters. The physical meaning and distances would be about the same because 1 yd is comparable with 1 m.

**57. **\(\vec{R}\) = −3 \(\hat{i}\) − 16 \(\hat{j}\)

**59. **\(\vec{E}\) = E \(\hat{E}\), E_{x} = + 178.9 V/m , E_{y} = −357.8 V/m, E_{z} = 0.0 V/m, \(\theta_{E}\) = −tan^{−1}(2)

**61.** a. \(\vec{R}_{B}\) = (12.278 \(\hat{i}\) + 7.089 \(\hat{j}\) + 2.500 \(\hat{k}\))km, \(\vec{R}_{D}\) = (−0.262 \(\hat{i}\) + 3.000 \(\hat{k}\))km

b. |\(\vec{R}_{B} − \vec{R}_{D}\)| = 14.414 km

**63.** a. 8.66

b. 10.39

c. 0.866

d. 17.32

**65. **\(\theta_{i}\) = 64.12°, \(\theta_{j}\) = 150.79°, \(\theta_{k}\) = 77.39°

**67.** a. −119.98 \(\hat{k}\)

b. 0 \(\hat{k}\)

c. +93.69 \(\hat{k}\)

d. −240.0 \(\hat{k}\)

e. +3.993 \(\hat{k}\)

f. −3.009 \(\hat{k}\)

g. +14.99 \(\hat{k}\)

h. 0

**69. **a. 0

b. 173,194

c. +199,993 \(\hat{k}\)

## Additional Problems

**71.** a. 18.4 km and 26.2 km

b. 31.5 km and 5.56 km

**73. **a. (r, \(\phi + \frac{\pi}{2}\))

b. (2r, \(\phi + 2 \pi\))

c. (3r, −\(\phi\))

**75. **d_{PM} = 33.12 nmi = 61.34 km, d_{NP} = 35.47 nmi = 65.69 km

**77.** proof

**79.** a. 10.00 m

b. 5\(\pi\) m, c. 0

**81. **22.2 km/h, 35.8° south of west

**83. **240.2 m, 2.2° south of west

**85. **\(\vec{B}\) = −4.0 \(\hat{i}\) + 3.0 \(\hat{j}\) or \(\vec{B}\) = 4.0 \(\hat{i}\) − 3.0 \(\hat{j}\)

**87. **proof

## Challenge Problems

**89. **G_{\(\perp\)} = 2375\(\sqrt{17}\) ≈ 9792

**91.** proof

## Contributors

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).