2 vectors orthogonal to 3 4-d vectors

[Damascus Road]

Greetings,
I'm a little stuck on this algebra question

Find 2 unit vectors orthogonal to v1 = <3,1,1,-1>, v2 = <-1,2,2,0> and v3 = <1,0,2,-1>

I know that this means that if I let z be the vector orthogonal to these 3 then,

z.v1 = z.v2= z.v3 = 0

And that my two vectors is likely +/- z.

I can do something like this:

z = <a,b,c,d>

z.v1 = z.v2

3a + b + c -d = -a + 2b + 2c

but now what? Or is this the wrong idea?

Thanks in advance!

 

[noumed]

This is rather tricky. Hm. If you're only using:
$$Z \cdot V1 = Z \cdot V2$$
then you're only solving for a vector that is orthogonal to V1 and V2, means you're excluding V3. So you need to use the third equality as well.

Even so, you have 4 unknowns and 3 equations. Something's amiss.

 

[tiny-tim]

Find 2 unit vectors orthogonal to v1 = <3,1,1,-1>, v2 = <-1,2,2,0> and v3 = <1,0,2,-1>

I know that this means that if I let z be the vector orthogonal to these 3 then,

z.v1 = z.v2= z.v3 = 0

And that my two vectors is likely +/- z.
…
z.v1 = z.v2

3a + b + c -d = -a + 2b + 2c

Greetings Damascus Road!

Yes, z.v1 = z.v2= z.v3 = 0 …

so your last line should be 3a + b + c -d = -a + 2b + 2c = 0 (= a + 2c - d) … three equations in four variables (which is ok, since you only need the ratios )

 

[Damascus Road]

Thanks for the replies!

I'm confused about the ratios however. Can I set the individual components equal?
3a = a = -a?

 

[tiny-tim]

Thanks for the replies!

I'm confused about the ratios however. Can I set the individual components equal?
3a = a = -a?

Nooo … these are three separate linear equations …

if it makes you happier, you can reduce the number of variables to 3 (a/d b/d and c/d) by rewriting the equations as 3a/d + b/d + c/d = 1 etc.