Recognitions:
Gold Member

## Finding perpendicular vector

1. The problem statement, all variables and given/known data

Find a unit vector in the xy plane which is perpendicular to A = (3,5,1).

2. Relevant equations

$$A_x{B_{x}} + A_y{B_{y}} + A_z{B_{z}} = \textbf{A} \cdot \textbf{B}\\\hat{\textbf{A}} = \frac{\textbf{A}}{|\textbf{A}|}$$

3. The attempt at a solution

In order to be perpendicular, AB = 0 since a perpendicular 90 degrees would mean cos(90) = 0, so the entire dot product becomes 0.

So:

$$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}$$

But since B doesn't exist on the z plane:

$$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}$$

So:

$$0 = 3{B_{x}} + 5{B_{y}}$$

Not sure what to do from here. Using:

$$\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}$$

How would I turn this B vector into a unit vector?

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Recognitions:
Homework Help
 Quote by PhizKid 1. The problem statement, all variables and given/known data Find a unit vector in the xy plane which is perpendicular to A = (3,5,1). 2. Relevant equations $$A_x{B_{x}} + A_y{B_{y}} + A_z{B_{z}} = \textbf{A} \cdot \textbf{B}\\\hat{\textbf{A}} = \frac{\textbf{A}}{|\textbf{A}|}$$ 3. The attempt at a solution In order to be perpendicular, A•B = 0 since a perpendicular 90 degrees would mean cos(90) = 0, so the entire dot product becomes 0. So: $$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}$$ But since B doesn't exist on the z plane: $$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}$$ So: $$0 = 3{B_{x}} + 5{B_{y}}$$ Not sure what to do from here. Using: $$\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}$$ How would I turn this B vector into a unit vector?
How do you get |B| from the components?

ehild

 you can take (1/√34)(-5i+3j) as your unit vector. edit-or also 5i-3j in place of -5i+3j.

Recognitions:
Gold Member

## Finding perpendicular vector

 Quote by PhizKid 1. The problem statement, all variables and given/known data Find a unit vector in the xy plane which is perpendicular to A = (3,5,1). 2. Relevant equations $$A_x{B_{x}} + A_y{B_{y}} + A_z{B_{z}} = \textbf{A} \cdot \textbf{B}\\\hat{\textbf{A}} = \frac{\textbf{A}}{|\textbf{A}|}$$ 3. The attempt at a solution In order to be perpendicular, A•B = 0 since a perpendicular 90 degrees would mean cos(90) = 0, so the entire dot product becomes 0. So: $$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}$$ But since B doesn't exist on the z plane: $$\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}$$ So: $$0 = 3{B_{x}} + 5{B_{y}}$$ Not sure what to do from here. Using: $$\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}$$ How would I turn this B vector into a unit vector?
If B is a unit vector, then B dotted with B has to be 1. What does this mean in component form?