# Help me determine a vector that's perpendicular (with work)

1. Sep 7, 2015

Problem: The vectors A and B together define a plane surface. Determine a vector of magnitude 3 that is perpendicular to that surface.

What I am thinking is Vector A is on the x-axis and B is on the y-axis. Therefore, the vector that I must determine, runs along the z-axis which is 90 degrees to that plain.

So, can I say that the answer is: The vector that must be determined is 90 degrees z-axis of x-axis with a magnitude of 3?

Sorry if what I am doing or saying is not even remotely close to the valid answer.

Can you please tell me how to approach this problem, maybe even go through it?

Here is a picture of what I think the Determined Vector of magnitude 3 is:
Picture URL: http://tinypic.com/r/25u16jd/8
http://tinypic.com/r/25u16jd/8

2. Sep 7, 2015

### fzero

Have you studied the cross product of vectors?

3. Sep 8, 2015

### gracy

Two vectors are perpendicular if their dot product is zero.

4. Sep 8, 2015

No I haven't, I am still learning the basics. I will spend the rest of tonight looking over it.

5. Sep 8, 2015

So |A|*|B|*cos(theta) = 0
cos(theta) = 0 / (|A|*|B|)
theta = arccos(0)
theta = 90 degress
I can say that the VECTOR that is perpendicular to the plane is 90 degrees counterclockwise and have its magnitude = 3?

6. Sep 8, 2015

### gracy

I think you should rather look at cross product

7. Sep 8, 2015

LOL, yeah I am, thank you.

8. Sep 8, 2015

### gracy

9. Sep 8, 2015

So I am not sure, but this is what I got.

First I find the cross product of A and B, since it will give me a vector that is perpendicular to both A and B.

A x B = C

And now that I have C in vector components, but from here I am lost. Do I multiply it by 3 since it wants a "vector with a magnitude of 3"?

so 3C = 3*(# i + #j + #k) is my answer?

Can you tell me if this is wrong and explain to me the correct way?

10. Sep 8, 2015

So I am not sure, but this is what I got.

First I find the cross product of A and B, since it will give me a vector that is perpendicular to both A and B.

A x B = C

And now that I have C in vector components, but from here I am lost. Do I multiply it by 3 since it wants a "vector with a magnitude of 3"?

so 3C = 3*[#i + #j + #k] is my answer?

Can you tell me if this is wrong and explain to me the correct way?

11. Sep 8, 2015

### fzero

You should compute the magnitude of $\mathbf{C} = \mathbf{A}\times \mathbf{B}$. You can either multiply $\mathbf{C}$ by an appropriate number to obtain a vector with magnitude 3 or derive a unit vector $\hat{\mathbf{C}}$ as an intermediate step and then multply that.

12. Sep 8, 2015

### RUber

This is just about right. However, as Fzero just said, if the magnitude of C is not 1, then 3*C will not have magnitude 3, it will have magnitude 3|C|.

13. Sep 8, 2015

So for the second method that you mentioned you mean like this?

A: 4i + 3j - 2k
B: 2i - 3j - 2k

AxB = C = i -12j - 18k

$\hat{\mathbf{C}}$ = C / |C|

3*$\hat{\mathbf{C}}$ = 3 * [-12i - 18k / sqrt(468)]

14. Sep 8, 2015

### fzero

Yes, that's the general idea. But with $\mathbf{A}$ and $\mathbf{B}$ as given I find
$$\mathbf{A}\times\mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & -2 \\ 2 & -3 & -2 \end{vmatrix}$$
gives a slightly different result from yours. So recheck your calculation before working out $\hat{\mathbf{C}}$.

15. Sep 8, 2015

### RUber

Where did your i component go?

16. Sep 8, 2015

Sorry for line "AxB = C = i -12j - 18k", I meant to say AxB = C = 0i -12j - 18k.

I forgot to go back and fix it after realizing my Num Lock was off.

17. Sep 8, 2015

I am not going to lie, I find physics and things associated with the course to be very hard and that I am struggling. Therefore, thank you everyone for you guys' help.

18. Sep 8, 2015

### RUber

I don't think you are handling the negatives properly.
I also get a different result.

19. Sep 8, 2015