Orthogonal Vectors (1 Viewer)

[davidgoes4wce]

any ideas on how to do this one?

 

[Shadowdude]

dot product, cross product

 

[davidgoes4wce]

I understand the dot product is 0. And I can do the cross product. I ended up with parameters in row reduced echelon augmented matrix.

 

[awesome-0_4000]

I understand the dot product is 0. And I can do the cross product. I ended up with parameters in row reduced echelon augmented matrix.

Why do you have parameters? The cross product should just give you a vector. Also you shouldn't need reduced row echelon form or augmented matrices, it's a determinant calculation for the cross product: Given two vectors A=(a1,a2,a3) and B=(b1,b2,b3) in R^3,
AxB=det((e1,e2,e3);(a1,a2,a3);(b1,b2,b3))
Where e1, e2 and e3 represent the standard basis vectors for R^3 and ";" denotes a new row.

 

[Shadowdude]

I understand the dot product is 0. And I can do the cross product. I ended up with parameters in row reduced echelon augmented matrix.

Are you using a 3x3 matrix where the first row is i, j, k and the second and third rows are the two input vectors to calculate the cross product?

 

[Silly Sausage]

You don't have to use matrices to do the cross product.
If you have vectors u=a+b+c and v=i+j+k.

You can do u x v =<bk-cj,ci-ak,aj-bi>

Using this to evaluate the cross product you should get u x v = 5i-4j+k.

 

[psyc1011]

If you don't know cross product then understanding dot products can help to figure out another way (which is what you probably did):

(a,b,c) * (1,2,3) = 0

(a,b,c,) * (-1,-1,1) = 0

where (a,b,c) is perpendicular to both specified vectors.

That gives this matrix,

.1 2 3 | 0
-1-1 1 | 0

Essentially, we are finding the kernel of this matrix. Row reduce,

1 2 3
0 1 4

1 0 -5
0 1 4

Thus the kernel of this matrix is span{(5,-4,1)} by inspection

 

[Shadowdude]

If you don't know cross product then understanding dot products can help to figure out another way:

(a,b,c) * (1,2,3) = 0

(a,b,c,) * (-1,-1,1) = 0

where (a,b,c) is perpendicular to both specified vectors.

That gives this matrix,

.1 2 3 | 0
-1-1 1 | 0

Essentially, we are finding the kernel of this matrix. Row reduce,

1 2 3 | 0
0 1 4 | 0

1 0 -5 | 0
0 1 4 | 0

Thus the kernel of this matrix is span{(5,0,1),(0,4,-1)}

I'm stuck... anyone pls continue?

Yeah that seems fine as a method, because the cross product vector isn't unique anyway

But your span doesn't seem right

 

[psyc1011]

Yeah that seems fine as a method, because the cross product vector isn't unique anyway

But your span doesn't seem right

FIxed

 

[Drongoski]

Are you using a 3x3 matrix where the first row is i, j, k and the second and third rows are the two input vectors to calculate the cross product?

Yes. This is very much the standard way to work out the cross-product, via the associated determinant (not strictly one). Then you derive the unit vectors of the 3 orthogonal vector to form the orthonormal basis.

 

[davidgoes4wce]

What does it mean by 'normalise a vector' . I did manage to find vector 'c' using the cross product initially.


From reading online sources, 'normalising a vector', you have to get the vector and divide it by the magnitude (in this case convert to a unit vector of length 1)

=a/|a|

 

[Drongoski]

You 'normalise' a vector to find the unit vector (i.e. vector of length 1) having the same direction. Thus your vector (1, 2, 3) = i + 2j + 3k has length sqrt(1^2 + 2^2 + 3^2) = sqrt(14); so you normalise it to:

 

[Carrotsticks]

What does it mean by 'normalise a vector' . I did manage to find vector 'c' using the cross product initially.


From reading online sources, 'normalising a vector', you have to get the vector and divide it by the magnitude (in this case convert to a unit vector of length 1)

=a/|a|

That is exactly right.

What you are essentially doing is constructing a 'new xyz plane'.

However, this new xyz plane needs units of course, much like how you have ijk.

You have your x, y and z 'axes' (the two given vector, and the vector you acquired upon taking the cross product), you just need to make those vectors have unit length to turn them into basis vectors, similarly to how ijk have unit length.

 

[davidgoes4wce]

It's the first time I have heard of 'normalising a vector'/



This was my work 'normalizing' of each of the 3 vectors. Could someone confirm if I was on the right track?

 

[awesome-0_4000]

It's the first time I have heard of 'normalising a vector'/



This was my work 'normalizing' of each of the 3 vectors. Could someone confirm if I was on the right track?

Yes, those expressions are correct. To check that they are unit vectors (ie normalised), compute the norm of the new vectors. If it is 1, then they are normalised.