Find vector C, using 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.

 

[tomcue]

To find vector C, we can use the fact that the dot product of perpendicular vectors is equal to zero. This means that the components of vector C must satisfy the equation 9.6C_i + 6.7C_j = 0. We can also use the given information about the dot product of A and C to form another equation: -4.8C_i + 6.8C_j = 20. We now have two equations with two unknowns (C_i and C_j). We can solve this system of equations using algebra or substitution to find the values of C_i and C_j. Once we have these values, we can use them to construct vector C, which will lie in the xy plane and be perpendicular to vector B.