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

[tex]
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}|}
[/tex]

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:

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}[/tex]

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

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}[/tex]

So:

[tex]0 = 3{B_{x}} + 5{B_{y}}[/tex]

Not sure what to do from here. Using:

[tex]\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}[/tex]

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

[tex]
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}|}
[/tex]

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:

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}[/tex]

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

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}[/tex]

So:

[tex]0 = 3{B_{x}} + 5{B_{y}}[/tex]

Not sure what to do from here. Using:

[tex]\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}[/tex]

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

[tex]
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}|}
[/tex]

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:

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}} + 1{B_{z}}[/tex]

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

[tex]\textbf{A} \cdot \textbf{B} = 3{B_{x}} + 5{B_{y}}[/tex]

So:

[tex]0 = 3{B_{x}} + 5{B_{y}}[/tex]

Not sure what to do from here. Using:

[tex]\hat{\textbf{B}} = \frac{\textbf{B}}{|\textbf{B}|}[/tex]

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?

 

[Nickie]

To find a unit vector that is perpendicular to A, we can use the cross product. The cross product of two vectors, A and B, is a vector that is perpendicular to both A and B. So, in this case, we can find a vector that is perpendicular to A by taking the cross product of A and any other vector in the xy plane.

Let's choose the vector (1,0,0) as our other vector. Then, the cross product of A and (1,0,0) would be:

\textbf{A} \times (1,0,0) = (0, -1, 5)

This vector is perpendicular to A, but it is not a unit vector. To make it a unit vector, we can divide it by its magnitude:

\hat{\textbf{B}} = \frac{(0, -1, 5)}{\sqrt{0^2 + (-1)^2 + 5^2}} = (0, -\frac{1}{\sqrt{26}}, \frac{5}{\sqrt{26}})

This is the unit vector in the xy plane that is perpendicular to A.