How to Determine Vector C Using Vectors A and B?

  • Thread starter Thread starter warfreak131
  • Start date Start date
  • Tags Tags
    Vector
Click For Summary
SUMMARY

The discussion focuses on determining vector C using vectors A and B in a two-dimensional plane. Given A = -4.8i + 6.8j and B = 9.6i + 6.7j, vector C must be perpendicular to B and satisfy the dot product condition with A equal to 20.0. The equations derived from the dot product conditions are 9.6Ci + 6.7Cj = 0 and -4.8Ci + 6.8Cj = 20. Solving these equations through substitution will yield the components of vector C.

PREREQUISITES
  • Understanding of vector operations, including dot products
  • Familiarity with trigonometric functions, specifically arctangent
  • Basic algebraic skills for solving simultaneous equations
  • Knowledge of vector representation in the Cartesian coordinate system
NEXT STEPS
  • Practice solving vector equations involving dot products
  • Explore the properties of perpendicular vectors in two dimensions
  • Learn about vector projections and their applications
  • Investigate the geometric interpretation of vectors in the xy-plane
USEFUL FOR

Students studying physics or mathematics, particularly those focusing on vector analysis and applications in two-dimensional space.

warfreak131
Messages
186
Reaction score
0

Homework Statement



Given vectors [tex]\bold{\vec{A}}=-4.8\bold{\hat{i}}+6.8\bold{\hat{j}}[/tex] and [tex]\bold{\vec{B}}=9.6\bold{\hat{i}}+6.7\bold{\hat{j}}[/tex], determine the vector [tex]\bold{\vec{C}}[/tex] that lies in the xy plane perpendicular to [tex]\bold{\vec{B}}[/tex] whose dot product with [tex]\bold{\vec{A}}[/tex] is 20.0

Homework Equations



[tex]\theta_{\vec{B}}=\arctan{\frac{6.7}{9.6}}=35 degrees[/tex]

Dot prod. of perpendicular vectors = 0, therefore
[tex]\vec{B}{\cdot}\vec{C}=B_{i}C_{i}+B_{j}C_{j}=0[/tex]
[tex]\vec{B}{\cdot}\vec{C}=9.6C_{i}+6.7C_{j}=0[/tex]

[tex]\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=20[/tex]
[tex]\vec{A}{\cdot}\vec{C}=-4.8C_{i}+6.8C_{j}=20[/tex]

The Attempt at a Solution



I'm not sure where to start, that's why I'm here :)
 
Physics news on Phys.org
It looks like you already have your 2 independent equations set up to solve for your 2 unknowns Ci and Cj. Just solve by a quick substitution and you will have vector C's i and j components. Its k component is 0 since it lies in the xy plane.
 

Similar threads

Replies
26
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
1
Views
2K