Solving Vector C: 3.00A - 4.00B

The final answer is 12.01i+14.94j.If you want a little more of a "proof" try looking at the components of the final vector. First you would have the i components of the first vector plus the i components of the second vector. So you have 3.00(1.23)i + 4.00(-2.08)i. Then you have the j components of the first vector plus the j components of the second vector. So you have 3.00(3.38)j + 4.00(-1.20)j. When you combine these two expressions you end up with 12.01i+14.94j.
  • #1
Heat
273
0
http://img523.imageshack.us/img523/4889/yf0148vj7.jpg​

Homework Statement



Use unit vectors to express the vector C, where C = 3.00A - 4.00B

Keep in mind that C, A, and B are vectors, I could not get the Latex to work.

Homework Equations



There are no relevant equations, but I don't know if the question above the one I am asking refers to this, as I got that:

Vector A = 1.23 i + 3.38j
Vector B = -2.08i - 1.2j

The Attempt at a Solution



My attempt would be to subtract -2.08 from 1.23 leading the final i to be 3.31 (as it's subtracting and negative with negative = +.) and the final j to be 4.58.

If this is how it is suppose to go, then do I plug those numbers in for A and B, multiply time their coeffiecients

ex: 3.00 (3.31) - 4.00 (4.58) = C ?

:confused:
 
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  • #2
Heat said:
My attempt would be to subtract -2.08 from 1.23 leading the final i to be 3.31 (as it's subtracting and negative with negative = +.) and the final j to be 4.58.

If this is how it is suppose to go, then do I plug those numbers in for A and B, multiply time their coeffiecients

ex: 3.00 (3.31) - 4.00 (4.58) = C ?

:confused:

No. What you're attempting to do is to multiply the differences of the i and j components by 3 and 4, respectively. What you actually need to do is multiply the vector A by 3, multiply the vector B by 4 and then take their difference. Multiplying a vector by a scalar is equivalent to multiplying each component of the vector by that scalar.
 
  • #3
*EDIT* deleted solution
You were solving for [tex]3.00(\vec{C}\hat{i}) + 4.00(\vec{C}\hat{j})[/tex]

But it asks for [tex]3.00\vec{A} - 4.00\vec{B}[/tex] or
[tex]3.00\vec{A}\hat{i} + 3.00\vec{A}\hat{j} - (4.00\vec{B}\hat{i} + 4.00\vec{B}\hat{j})[/tex]
 
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  • #4
odie5533, as per the forum's guidelines, you are not supposed to reveal the solution to the problems posted in the homework help forums. At most, you can guide the person to the solution. Apart from being a rule, it is also good practice to not provide the solution, since by copying the answer the poster does not gain anything apart from a few points for a single question.
 
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  • #5
Sorry neutrino. Fixed my response.
 
  • #6
@ odie5533 Don't worry about me looking at the solution, I logged in too late:rolleyes: , but I thank you for the explanation.

[tex]3.00\vec{A}\hat{i} + 3.00\vec{A}\hat{j} - (4.00\vec{B}\hat{i} + 4.00\vec{B}\hat{j})[/tex]


Here is what I did, I just don't know where it goes from here.

[3.00 (1.23 i) + 3.00 ( 3.38j)] - [(4(-2.08i) + 4(-1.2j))] =

3.69i + 10.14j + 8.32i + 4.8j

Do I add the like terms resulting into:

12.01i + 14.94j

:uhh:

PS: I'm here to learn, as I would need to know all of this later on.:approve:
 
  • #7
Yes, Heat. Exactly how you have done it is right. First multiplying the components by the constant, then performing the vector subtraction.
 

What is Vector C?

Vector C is a mathematical term used to represent a quantity that has both magnitude and direction. It is typically denoted by a boldface letter, such as C.

What does the equation 3.00A - 4.00B represent?

The equation 3.00A - 4.00B is a specific way of representing Vector C. The numbers 3.00 and 4.00 represent the magnitudes of two different vectors, A and B, while the minus sign indicates that they are subtracted from each other. The resulting value will be the magnitude of Vector C.

How do you solve for Vector C when given 3.00A and 4.00B?

To solve for Vector C, you will need to use the Pythagorean theorem and trigonometric functions. First, you will need to find the magnitude of Vector C by taking the square root of (3.00^2 + 4.00^2). Then, you will need to use trigonometric functions to find the direction of Vector C.

What are some real-life applications of solving Vector C?

Solving Vector C is commonly used in physics, engineering, and navigation. It can be used to calculate the net force acting on an object, determine the direction and magnitude of a current in an electrical circuit, or find the resultant velocity of an airplane given its wind velocity and airspeed.

What are some common mistakes when solving for Vector C?

One common mistake is forgetting to take the square root when finding the magnitude of Vector C. Another mistake is using the wrong trigonometric function to find the direction of Vector C. It is also important to ensure that the units of all vectors used in the equation are consistent.

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