Solving Mutually Perpendicular Vector Components

In summary, CinderBlockFist found a system of 3 equations in 3 unknowns that when multiplied by 2 and 3 give 6Ux+ 6Vy+ 12=0 and 6+ 4Vy+ 3Wz=0.
  • #1
CinderBlockFist
86
0
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.
 
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  • #2
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
 
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  • #3
what is cramer, what do i substitute?
 
  • #4
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.
 
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  • #5
Cinder are you familiar with System fo equations?, because i see no problem with Final's answer...
 
  • #6
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
 
  • #7
CinderBlockFist said:
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.
 
  • #8
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
 
  • #9
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.
 
  • #10
CinderBlockFist said:
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
 
  • #11
CinderBlockFist said:
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.
 
  • #12
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.
 
  • #13
HallsofIvy said:
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 =)
 
  • #14
Ah, yes! The ever-popular "fight with one hand tied behind my back method"!
 

1. What is a vector component problem?

A vector component problem is a type of mathematical problem that involves breaking down a vector into its horizontal and vertical components in order to solve for unknown values.

2. How do you find the horizontal and vertical components of a vector?

To find the horizontal and vertical components of a vector, you can use trigonometric functions such as sine, cosine, and tangent. The horizontal component can be found using cosine and the vertical component can be found using sine.

3. What is the purpose of solving vector component problems?

The purpose of solving vector component problems is to understand the individual components of a vector and how they contribute to its overall magnitude and direction. This can be useful in various fields such as physics, engineering, and navigation.

4. What are some real-life applications of vector component problems?

Vector component problems are used in many real-life applications, such as determining the force and direction of wind on a sailboat, calculating the trajectory of a projectile, and analyzing the forces acting on a structure.

5. Are there any tips for solving vector component problems?

One tip for solving vector component problems is to draw a diagram to visualize the problem and identify the horizontal and vertical components. It can also be helpful to break the vector into its components early on in the problem, rather than trying to solve for the overall magnitude and direction first.

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