How to Find a Unit Vector Perpendicular to Another in the XY Plane?

[PhizKid]

Homework Statement

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

Homework Equations

<br /> 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}|}<br />

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?

 

[ehild]

Homework Statement

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

Homework Equations

<br /> 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}|}<br />

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

 

[andrien]

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

 

[Chestermiller]

Homework Statement

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

Homework Equations

<br /> 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}|}<br />

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?