Find components of vector C from vectors A and B

  • #1

Homework Statement



Given vectors [tex]\vec{A} = 5.0\hat{i} - 6.5\hat{j}[/tex] and [tex]\vec{B} = -3.5\hat{i}= 7.0\hat{j}[/tex]. Vector [tex]\vec{C}[/tex] lies in the xy-plane. Vector [tex]\vec{C}[/tex] is perpendicular to [tex]\vec{A}[/tex] and the scalar product of [tex]\vec{C}[/tex] with [tex]\vec{B}[/tex] is 15.0. Find the vector components of [tex]\vec{C}[/tex].

Homework Equations



[tex]\vec{A}{\cdot}\vec{C} = 0 [/tex]
[tex]\vec{B}{\cdot}\vec{C} = 15 [/tex]

[tex]\vec{B}{\cdot}\vec{C}=B_{i}C_{i}+B_{j}C_{j}=15 [/tex]
[tex]\vec{B}{\cdot}\vec{C}=-3.5C_{i}+7.0C_{j}=15[/tex]

[tex]\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=0[/tex]


The Attempt at a Solution



Since the vectors A and C are perpendicular
[tex]\vec{A}{\cdot}\vec{C} = 0 [/tex]
Then,
[tex]\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=0[/tex]
[tex]\vec{A}{\cdot}\vec{C}=5.0_{i}C_{i}-6.5_{j}C_{j}=0[/tex]
[tex]C_{j}=\frac{5.0_{i}C{i}}{6.5}[/tex]

Plug in [tex]C_{j}[/tex] into the other scalar equation and solve for [tex]C_{i}[/tex]. Basic substitution. However I keep getting the wrong answer. Am I approaching the problem incorrectly or is my algebra wrong?

The correct answer is [tex]C_{x} = 8.0[/tex] and [tex]C_{y} = 6.1[/tex]
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
Hi casemeister06, welcome to PF.
-3.5Ci + 7Cj = 15.......(1)
5.0Ci - 6.5Cj = 0.........(2)
Multiply by 0.7 to eq. (2) and add it to eq.(1) and solve for Cj.
 
  • #3
Yeah, I don't know what I was doing, but I got it right now. I think I was messing up on my algebra or something. Thanks for the help.
 

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