Finding Orthogonal Vectors in R4 with Norm 1

[sevag00]

Homework Statement

Find two vectors in R4 of norm 1 that are orthogonal to the vectors u = (2, 1, −4, 0),
v = (−1, −1, 2, 2) and w = (3, 2, 5, 4).

Homework Equations

The Attempt at a Solution

What i did was, i let a vector x = (x1, x2, x3, x4) that has a norm of 1 and orthogonal to u,v and w
i.e. x.u = 0; x.v = 0 and x.w = 0.
This where I'm stuck. Some help would be appreciated. Thanks.

 

[tiny-tim]

hi sevag00!

(try using the X₂ button just above the Reply box )

hint: how would you find a vector (x₁,x₂,x₃) orthogonal to (a,b,c) and (d,e,f) ?

 

[sevag00]

Using dot product. x₁.a + x₂.b + x₃.c = 0. Same for (d, e, f).

 

[tiny-tim]

how about the cross product ?

 

[sevag00]

Yeah, but the chapter only explains the dot product. So, i think the problem should be solved using dot product.

 

[tiny-tim]

ok, in that case you have three simultaneous equations (from the dot products) …

solve them the usual way ​

 

[sevag00]

3 equations with 4 unknowns. Will this yield a solution?

 

[tiny-tim]

3 equations with 3 unknown ratios

so yes, you can find the ratios ​

 

[sevag00]

Ah yes. 4 equations with 4 unknowns.

 

[HallsofIvy]

how about the cross product ?

Yeah, but the chapter only explains the dot product. So, i think the problem should be solved using dot product.

Further, the cross product is defined only in three dimensions while this problem is in four dimensions.

 

[sevag00]

Wait a minute. You got me confused. There are 3 equations with 4 unknown. The equation of the norm cannot be included with the other three.

 

[tiny-tim]

Further, the cross product is defined only in three dimensions …

i was talking about three dimensions

There are 3 equations with 4 unknown. The equation of the norm cannot be included with the other three.

there are only three unknown ratios

so you can find the ratios​

 

[sevag00]

What ratios? I want to find the components of x.

 

[tiny-tim]

x₂/x₁, x₃/x₁, x₄/x₁

 

[sevag00]

Not getting your approach.

 

[tiny-tim]

divide your three dot-product=0 equations by x₁

that gives you three equations in x₂/x₁, x₃/x₁, and x₄/x₁

then solve them ​

 

[sevag00]

I think i did it right. I replaced x₂/x₁ with X x₃/x₁ with Y and x₄/x₁ with Z. Then solved 3 equations with 3 unknowns.

 

[Ray Vickson]

Homework Statement

Find two vectors in R4 of norm 1 that are orthogonal to the vectors u = (2, 1, −4, 0),
v = (−1, −1, 2, 2) and w = (3, 2, 5, 4).

Homework Equations

The Attempt at a Solution

What i did was, i let a vector x = (x1, x2, x3, x4) that has a norm of 1 and orthogonal to u,v and w
i.e. x.u = 0; x.v = 0 and x.w = 0.
This where I'm stuck. Some help would be appreciated. Thanks.

You can write the three equations in four unknowns as three equations in three unknowns but with the fourth unknown on the right-hand-side. In other words, you can express three of the x_i as functions of the fourth. For example, you can write the second equation $-1 x_1 - 1 x_2 + 2 x_3 + 2 x_4 = 0$ as $-1 x_1 - 1 x_2 + 2 x_3 = -2 x_4$, or $x_1 + x_2 - 2x_3 = 2x_4$. Do the same thing for the third equation; the first already has the right form. Now you can solve for $x_1, x_2, x_3$; this will give them as formulas involving $x_4$.

BTW: the method above is a standard way of dealing with such problems; it is used all the time in such fields as linear optimization (linear programming).

 

[sevag00]

I know that method. But then those equations will become parameters. They will depend on the value of x₄ which has infinite solutions.

 

[Ray Vickson]

I know that method. But then those equations will become parameters. They will depend on the value of x₄ which has infinite solutions.

Of course; that is the whole point. So, if you want to find two different solutions, just give two different values for $x_4$---any two values you like, except 0. (Putting $x_4 = 0$ gives the zero vector, which cannot be normalized to length = 1!)

 

[sevag00]

I want direct solution for the x-es. I'm not going to plug in millions of values to see if the norm is 1.

 

[tiny-tim]

put x₄ = 1, then normalise it

 

[Ray Vickson]

I want direct solution for the x-es. I'm not going to plug in millions of values to see if the norm is 1.

You have completely misunderstood what some of us have been saying. First find two vectors that satisfy the perpendicularity conditions. THEN normalize them.

For example, if I start with a vector like $v = (1,2,3,4)$ I can get a normalized version as
u = \frac{v}{\text{norm of }v}
If v was perpendicular to some vector w then so is u, and u has norm 1. That's all there is to it.