How Do You Find a Unit Vector Orthogonal to Two Given Vectors?

[kwal0203]

Homework Statement

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

Any help appreciated thanks!

 

[haruspex]

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

 

[kwal0203]

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

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

 

[haruspex]

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

 

[kwal0203]

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

I could try that, how do I do it?

 

[haruspex]

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

 

[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?

 

[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.

 

[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?

 

[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.

 

[kwal0203]

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

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

 

[kwal0203]

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

 

[Fredrik]

let x=(x,y,z)

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

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

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

x+y=z-x
y=z

something like this?

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 canceled x and -x from what you had left.