Proving vectors are orthogonal

[psycho81]

Homework Statement

first the question asks find the jacobian matrix of

(ucosv)
(usinv )
( w )
i have the matrix

( cos(v) , -usin(v) , 0)
( sin(v) , ucos(v) , 0)
( 0 , 0 , 1)

the question asks to show that the columns are orthogonal vectors, doing the dot product column 1 and 2 with 3 is easy because of the 0's, but how do I dot product the 1st two vectors?

Homework Equations

The Attempt at a Solution

(cosv x -usinv) + (sinv x ucosv) + (0 x 0) = 0

would just this suffice you think?

 

[psycho81]

thats F(u,v,w) at the start

 

[HallsofIvy]

Homework Statement

first the question asks find the jacobian matrix of

(ucosv)
(usinv )
( w )
i have the matrix

( cos(v) , -usin(v) , 0)
( sin(v) , ucos(v) , 0)
( 0 , 0 , 1)

the question asks to show that the columns are orthogonal vectors, doing the dot product column 1 and 2 with 3 is easy because of the 0's, but how do I dot product the 1st two vectors?

Homework Equations

The Attempt at a Solution

(cosv x -usinv) + (sinv x ucosv) + (0 x 0) = 0

would just this suffice you think?

Well, why not do the multiplication:
-u cos v sin v+ u cos v sin v= 0.

 

[psycho81]

thanks, now I have to calculate the corresponding unit vectors

would that just be

e_u= (sqrt( cos²v + sin²v+0)) = 1

e_v = (sqrt(-usin²v + ucos²v +0)) =1

e_w = (sqrt( 0 + 0 + 1²) = 1

or have i done this wrong?

 

[psycho81]

in order to be a unit vector

sqrt(cos²v+sin²v)=1

sqrt(v) = 1

so v must be 1?

am in on the right track?

 

[Inertigratus]

You don't put it equal to one. Secondly, a vector is a vector, a set of coordinates. Not a scalar.
A unit vector is a vector whose absolute value of the magnitude is equal to one, meaning the square root of the coordinates squared and then summed up should be 1.
For example (1, 1, 0). sqrt((1^2)+(1^2)+(0^2))=sqrt(2) and is not equal to 1.
Therefor to turn it into a unit vector you have to divide it by the absolute value of the magnitude, which here is sqrt(2).
So (1, 1, 0) is a vector. (1/sqrt(2))*(1, 1, 0) is a unit vector.
Hope you understand. Good luck!

 

[psycho81]

how can you give it an absolute value of the magnitude when you don't know what u, v, or w is though? or do you just set them to 1

 

[psycho81]

ahhh so it would be over sqrt(v)??

and sqrt(uv)

and the last one already has length 1?

please be right I am starting to lose the will to live.

 

[Inertigratus]

You have your matrix:
( cos(v) , -usin(v) , 0)
( sin(v) , ucos(v) , 0)
( 0 , 0 , 1)

Your vectors are (cos(v), sin(v), 0), (-usin(v), ucos(v), 0) and (0, 0, 1)

(0, 0, 1) is obviously already a unit vector.

sqrt(cos^2(v) + sin^2(v)) = sqrt(1) = 1
=> so (cos(v), sin(v), 0) is already a unit vector as well.

sqrt((-usin(v))^2 + (ucos(v))^2) = sqrt(u^2 * sin^2(v) + u^2 + cos^2(v))
= sqrt(u^2 * (sin^2(v) + cos^2(v)) = sqrt(u^2) = u

Therefor 1/u * (-usin(v), ucos(v), 0) is a unit vector.

You can test this yourself, since you already know that the magnitude is u that means that the magnitude of 1/u * (-usin(v), ucos(v), 0) is equal to u/u which is equal to 1.
That's the reason you have to divide by the length of the vector to get a unit vector.