How to Determine Vector C Using Vectors A and B?

[warfreak131]

Homework Statement

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

Homework Equations

\theta_{\vec{B}}=\arctan{\frac{6.7}{9.6}}=35 degrees

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

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

The Attempt at a Solution

I'm not sure where to start, that's why I'm here :)

 

[mplayer]

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.