# Finding orthogonal vector

1. Mar 6, 2013

### kwal0203

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

Find a unit vector that is orthogonal to both $u=(1,1,0)$ and $v=(-1,0,1)$

Any help appreciated thanks!

2. Mar 6, 2013

### haruspex

What do you know about the direction of the cross product of two vectors?

3. Mar 6, 2013

### kwal0203

Yeah, I see where you are going with that but I need to answer the question without using the cross product

4. Mar 6, 2013

### haruspex

Ok, so how about supposing the vector is (x, y, z) and obtaining some equations based on dot products?

5. Mar 6, 2013

### kwal0203

I could try that, how do I do it?

6. Mar 6, 2013

### haruspex

If you take the dot product with each given vector, what should the result be?

7. Mar 6, 2013

### kwal0203

$u\cdot v=(1,1,0)\cdot (-1,0,1)=-1+0+0=-1$

$u\cdot v=\left \| u \right \|\left \| v \right \|cos\theta$

the dot product represents the angle between the vectors, is that correct?

8. Mar 6, 2013

### haruspex

No, I meant take the dot product of (x, y, z) (a vector intended to be orthogonal to the two given vectors) with each of them.

9. Mar 6, 2013

### kwal0203

let $x=(x,y,z)$

$x\cdot u=(x,y,z)\cdot (1,1,0)$
$x+y=\left \| x \right \|\left \| u \right \|cos\theta=0$

$x\cdot v=(x,y,z)\cdot (-1,0,1)$
$z-x=\left \| x \right \|\left \| v \right \|cos\theta=0$

$x+y=z-x$
$y=z$

something like this?

Last edited: Mar 6, 2013
10. Mar 6, 2013

### haruspex

I don't understand what u and v represent in there. looks like in the first line they represent (x,y,z) and (1,1,0), and in the second line (x,y,z) and (-1,0,1). But then you deduce an equation based on u and v representing the same in both cases??
If two vectors are orthogonal, what is their dot product?
Edit: It's late here ... and so to bed.

11. Mar 6, 2013

### kwal0203

Whoops, I fixed up the code. Does it look better now?

The dot product of two orthogonal vectors is equal to zero.

12. Mar 6, 2013

### kwal0203

Ah, I figured it out... thanks for your help!

13. Mar 6, 2013

### Fredrik

Staff Emeritus
It's confusing to have one symbol mean two different things, so I recommend that you don't use this notation. I'll use $p=(x,y,z)$.

You should write this as
$$0 =p\cdot u=(x,y,z)\cdot (1,1,0) =x+y.$$ No need to mention the angle between the vectors. Also, you're confusing your readers when you put the zero at the end, instead of at the beginning next to $p\cdot u$ which is equal to 0 by definition of p. (Edit: OK, I see that the thing at the end is also obviously equal to zero because the angle between the vectors is $\pi/2$, but I still prefer to put the zero at the beginning, where no thought is required). You also left out an equality sign.

Similarly,
$$0=p\cdot v =(x,y,z)\cdot (-1,0,1) =-x+z.$$

Yes, something like that. But not that. First you threw away useful information from the equalities $p\cdot u=0$ and $p\cdot v=0$, and then you incorrectly cancelled x and -x from what you had left.

Last edited: Mar 6, 2013