If c = A· E, then A - cE = 0.

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In summary, the conversation discusses the use of unit vectors in calculating vector components. The use of the parallelogram law is mentioned to show that A - cE is perpendicular to E and cE. The conversation also touches on the concept of the scalar product and its relation to the perpendicularity of vectors.
  • #1
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Homework Statement



Let E be any unit vector, that is a vector of norm 1. Let c be the component
of A along E. We saw that c = A·E. Then A - cE is perpendicular to E, and A = A - cE + cE
Then A - cE is also perpendicular to cE ...

Homework Equations


The Attempt at a Solution



I just can't seem to figure where they got A - cE = 0 from. Everyone just shows A - cE = 0 is true on its own instead of showing if c = A·E, then A - cE = 0* which is what I think the author means by "We saw that c = A·E. Then A - cE is perpendicular to E".

Please, elaborate a little. Thanks.

edit: * if c = A·E, then A - cE · E = 0
 
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  • #2
Eh? Where does it say A - cE = 0? Are you trying to prove that A-cE is perpendicular to E?
 
  • #3
qspeechc said:
Eh? Where does it say A - cE = 0? Are you trying to prove that A-cE is perpendicular to E?

Oh yeah, A - cE . E = 0

My bad.
 
  • #4
You mean (A-cE)·E=0, I suppose. Geometrically, if E is a unit vector, then is cE not the vector component of A along E? So what is A-cE?

EDIT: heh, I wrote something silly, so I had to delete it
 
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  • #5
qspeechc said:
You mean (A-cE)·E=0, I suppose. Geometrically, if E is a unit vector, then is cE not the vector component of A along E? So what is A-cE?

My book has the picture of this situation. Geometrically it's clear that A-cE is a vector perpendicular to E and cE.

A = A - cE + cE by parallelogram law. But why do we need A = A - cE + cE to see that A-cE is also perpendicular to cE when (A-cE)· E=0 implies (A-cE)· cE=0?

qspeechc said:
(A-cE)·E = (A-A·E·E)·E

Right? What is E·E when E is a unit vector?

E·E = 1.
 
  • #6
Yeah, that second part I wrote is clearly rubbish, so I deleted it. The scalar product of three vectors is nonsense, because the product of two is a scalar.

A = A - cE + cE because -cE+cE=0. I think the point about A = A - cE + cE is that A = (A - cE) + cE, i.e. the vectors A, (A-cE) and cE form a right-angled triangle, with A as the hypotenuse. This is obvious geometrically, and I don't know what more is required.
 

Related to If c = A· E, then A - cE = 0.

1. What does the equation c = A· E mean?

The equation c = A· E means that the value of c is equal to the product of A and E. In other words, c is the result of multiplying A and E together.

2. What is the significance of the equation c = A· E?

The equation c = A· E is significant because it shows the relationship between c, A, and E. It tells us that c is dependent on the values of A and E, and that changes in either A or E will affect the value of c.

3. How can we interpret the equation c = A· E?

The equation c = A· E can be interpreted as representing a linear relationship between c, A, and E. It can also be interpreted as a special case of the more general formula c = A · B, where B is any other variable or constant.

4. What does the expression A - cE = 0 mean in relation to the equation c = A· E?

The expression A - cE = 0 means that the difference between A and cE is equal to 0. In other words, it means that A and cE are equal to each other. When this is true, it also means that c = A· E will also be true, since c is the product of A and E.

5. How can we use the equation c = A· E to solve for unknown values?

The equation c = A· E can be rearranged to solve for any of the variables. For example, if we know the values of c and E, we can solve for A by dividing both sides of the equation by E. Similarly, if we know the values of c and A, we can solve for E by dividing both sides of the equation by A. This equation can also be used to find the value of one variable if we know the values of the other two.

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