Ma 125 Exam 1 Solutions

1. [10] 2. [10] 3. [6] 4. [12] 5(a). [4] 5(b). [4] 5(c). [4] Total: [50]




Ma 125 Exam 1 Feb 6, 2023

Name: Solutions
Check your Section:
✷ A - Serbin (9:00am) ✷ B - Schwartz (9:00am) ✷ C - Serbin (10:00am)
✷ D - Monika (10:00am) ✷ E - Miller (11:00am) ✷ F - Monika (11:00am)
✷ G - Miller (1:00pm) ✷ H - Schwartz (2:00pm)

Closed book and closed notes. Answers must include supporting work.
Calculators and cell phones out of sight.

Pledge and Sign:




1. Two forces, represented by the vectors F 1 = 3i + 3j − k and F
 2 = 5i + j + 4k are
acting on a particle. The resultant force F moves the particle from the point
A(1, −2, 3) to the point B(2, −4, 5).
−→
(a) [4 pts] Find a unit vector in the direction of AB.
−→
(b) [4 pts] Compute the vector component of F parallel to the vector AB.
 in moving the particle (the distance is
(c) [2 pts] Find the work done by the force F
measured in meters and the force in newtons).

−→ −→ p √
Solution: (a) We have AB = ⟨1, −2, 2⟩, so ∥ AB∥ = 12 + (−2)2 + 22 = 9 = 3 and
−→
a unit vector in the direction of AB is
1 2 2

1 −→

· AB = ,− ,
3 3 3 3

 = F1 + F
(b) First for all, we have F 2 = 8i + 4j + 3k. The vector component of F
−→ −→
 onto AB.
parallel to the vector AB is the projection of F Hence,
−→  !  
) = AB · F −→ 1 · 8 + (−2) · 4 + 2·3
proj−AB
→ (F
−→ 2 · AB = · ⟨1, −2, 2⟩
∥ AB∥ 3 2


2 4 4
 
2
= · ⟨1, −2, 2⟩ = ,− ,
3 3 3 3

(c) We have
−→
W = AB · F = 1 · 8 + (−2) · 4 + 2 · 3 = 6 J