Solving Mutually Perpendicular Vector Components

[CinderBlockFist]

Guys, I am stuck on this problem. I have the answer from the back of the book, but I don't know how to solve it. I tried to solve it by isolating Ux, Vy, and trying to solve it by simultaneous equations, but I always end up getting something like Vy = Vy , or something like Vy - Vy = 0. It keeps going in circles, maybe this isn't the way to go. But anyways, this is the problem:



(note: the x, y , and z are written as small sub next to the U, V, and W, I just can't type it like it is in the book)

The three vectors

U = Uxi + 3j + 2k
V = -3i + Vyj + 3k
W = -2i + 4j + Wzk

are mutually perpendicular. Use the dot product to determine the compnents of Ux, Vy, and Wz.


The answers in the back of the book are Ux = 2.857, Vy = 0.857, Wz = -3.143.


I am using the dot product definition: U dot V = UxVx+UyVy+UzVz


so I get (Ux)(-3)+(3)(Vy)+(2)(3) = 0

I set it all to zero because they are perpendicular to each other.

So simplifying for Ux, i get Ux = Vy + 2. THen I plug Vy +2 for Ux, but then I go in circles from there.

 

[Final]

Using the dot product definition: U dot V = UxVx+UyVy+UzVz = |U||V|cos(a)
you have a system of 3 equation:

-3U + 3V + 6 = 0
6 + 4V +3W = 0
-2U + 12 + 2W = 0

You can solve it by Cramer or by substitution
The solution are: U =20/7 V=6/7 W=-22/7

 

[CinderBlockFist]

what is cramer, what do i substitute?

 

[CinderBlockFist]

What do I put for the |U||V|, since those are the magnitudes, I don't know how to find them because we are given only the components, and some are (Ux,Vy,Wz) are not actuallly given, so I can't compute it in the radical to get a value.

 

[Pyrrhus]

Cinder are you familiar with System fo equations?, because i see no problem with Final's answer...

 

[CinderBlockFist]

how did u get the -3, 4, and the 12, they are not given in the problem? They are only variables Ux, Vy, and Wz

 

[Pyrrhus]

how did u get the -3, 4, and the 12, they are not given in the problem? They are only variables Ux, Vy, and Wz

In Final's post U is Ux, V is Vy, and W is Wz.

 

[CinderBlockFist]

Yes, but in his 3 equations, he has -3U in the first one, and 6, in the second, and , man sorry i just don't see it

 

[CinderBlockFist]

can someone show me the steps to get those values, cause when i tried to isolate Ux, or Vy, or Wz, in a system, then i would get like Vy - Vy =0 or something useless.

 

[Pyrrhus]

Yes, but in his 3 equations, he has -3U in the first one, and 6, in the second, and , man sorry i just don't see it

The first equation is U dot V
The second equation is V dot W
The third equation is U dot W

 

[Pyrrhus]

can someone show me the steps to get those values, cause when i tried to isolate Ux, or Vy, or Wz, in a system, then i would get like Vy - Vy =0 or something useless.

There are 3 unknowns and 3 equations, this is a solvable system, check what you are doing.

 

[HallsofIvy]

No offense intended, CinderBlockFist, but this a lot like a guy asking a question about calculus and not being able to do arithmetic. If you are studying linear algebra, you certainly would be expect to be able to solve 3 linear equations in 3 unknowns. You may have had that so long ago you have forgotten.

The three vectors are
U = Uxi + 3j + 2k
V = -3i + Vyj + 3k
W = -2i + 4j + Wzk

U.V= -3Ux+ 3Vy+ 6= 0
U.W= -2Ux+ 12+ 2Wz= 0
W.V= 6+ 4Vy+ 3Wz= 0. Three linear equations in 3 unknowns.

There are many different ways to solve these but the simplest is probably:
Multiply the first equation by 2 to get -6Ux+ 6Vy+ 12= 0
Multiply the second equation by 3 to get -6Ux+ 36 +6Wz=0
Of course, the purpose of the multiplications was to get Ux with the same coefficient, -6, in both. If we subtract the second of those equations from the first the two "-6Ux" terms cancel and we get 6Vy+ 12- 36- 6Wz= 0 or 6Vy- 6Wz- 24= 0.

The last equation did not have Ux in it to begin with so now I have two equations in 2 unknowns: 6Vy- 6Wz- 24= 0 and 6+ 4Vy+ 3Wz= 0. If we divide the first of those by 2 we get 3Vy- 3Wz- 12=0 and adding that two the second will cause the "3Wz" terms to cancel: 3Vy- 3Wz- 12+ 6+ 4Vy+ 3Wz= 7Vy- 6= 0. That one equation in one unknown we can solve to get Vy= 6/7.

Putting that back into the original equation 6+ 4Vy+ 3Wz= 0, we get 6+ 24/7+ 3Wz= 0 so 3Wz= -6- 24/7= -66/7 so Wz= -22/7. Finally, we can put Vy= 6/7 into the original equation -3Ux+ 3Vy+ 6= 0 which is the same as Ux= Vy+ 2= 6/7+ 2= 20/7.

Putting all of those together, Ux= 20/7, Vy= 6/7, and Wz= -22/7 just as Final said originally.

 

[CinderBlockFist]

No offense intended, CinderBlockFist, but this a lot like a guy asking a question about calculus and not being able to do arithmetic. If you are studying linear algebra, you certainly would be expect to be able to solve 3 linear equations in 3 unknowns. You may have had that so long ago you have forgotten.

The three vectors are
U = Uxi + 3j + 2k
V = -3i + Vyj + 3k
W = -2i + 4j + Wzk

U.V= -3Ux+ 3Vy+ 6= 0
U.W= -2Ux+ 12+ 2Wz= 0
W.V= 6+ 4Vy+ 3Wz= 0. Three linear equations in 3 unknowns.

There are many different ways to solve these but the simplest is probably:
Multiply the first equation by 2 to get -6Ux+ 6Vy+ 12= 0
Multiply the second equation by 3 to get -6Ux+ 36 +6Wz=0
Of course, the purpose of the multiplications was to get Ux with the same coefficient, -6, in both. If we subtract the second of those equations from the first the two "-6Ux" terms cancel and we get 6Vy+ 12- 36- 6Wz= 0 or 6Vy- 6Wz- 24= 0.

The last equation did not have Ux in it to begin with so now I have two equations in 2 unknowns: 6Vy- 6Wz- 24= 0 and 6+ 4Vy+ 3Wz= 0. If we divide the first of those by 2 we get 3Vy- 3Wz- 12=0 and adding that two the second will cause the "3Wz" terms to cancel: 3Vy- 3Wz- 12+ 6+ 4Vy+ 3Wz= 7Vy- 6= 0. That one equation in one unknown we can solve to get Vy= 6/7.

Putting that back into the original equation 6+ 4Vy+ 3Wz= 0, we get 6+ 24/7+ 3Wz= 0 so 3Wz= -6- 24/7= -66/7 so Wz= -22/7. Finally, we can put Vy= 6/7 into the original equation -3Ux+ 3Vy+ 6= 0 which is the same as Ux= Vy+ 2= 6/7+ 2= 20/7.

Putting all of those together, Ux= 20/7, Vy= 6/7, and Wz= -22/7 just as Final said originally.

Ivy, THank you for the great response, yes I get it! I forgot, I was trying to solve using only 2 of the equations, no wonder i got no where. And I forgot about that technique to match of the coeff. Thanks again =)

 

[HallsofIvy]

Ah, yes! The ever-popular "fight with one hand tied behind my back method"!