- #1

- 7

- 0

## 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: