If c = A· E, then A - cE = 0.

[press]

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

 

[qspeechc]

Eh? Where does it say A - cE = 0? Are you trying to prove that A-cE is perpendicular to E?

 

[press]

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.

 

[qspeechc]

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

 

[press]

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?

(A-cE)·E = (A-A·E·E)·E

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

E·E = 1.

 

[qspeechc]

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.