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Topic review (newest first)
- Atled
- 2005-09-10 10:54:40
About myself...
25/m/Ca
EE student at UC Davis
hobbies: cycling, skating, surfing although there are no oceans near davis 
- almukmin
- 2005-09-09 08:19:40
I just got an answer from my teacher. a.a is basically |a|^2
- MathsIsFun
- 2005-09-09 07:17:22
So long ago since I did dot or cross products, but I may venture an answer:
Dot Product: X.Y = |X|*|Y| * cos(θ)
So: A.A = |A|*|A| * cos(θ) = |A|*|A| * 1 (because θ=0, ie there is a 0 angle between A and A) = 2*2 = 4
- almukmin
- 2005-09-08 22:59:49
Atled wrote:
A . (A + 3B +7C) = A.A + 3(A.B) + 7(A.C)
A.B and A.C are given
A . (A + 3B +7C) = A.A +(3*3) +(7*-1)
A . (A + 3B +7C) = A.A + 2
whats A.A
X.Y = |X|*|Y| * cos(θ)
A is normal to A there for θ = 0
or
The angle between A and A is 0
cos(0) = 1
A.A = 4
A . (A + 3B +7C) = 4 + 2 =6
hope that helps
I don't understand how come A.A is 4. Is |A| = A? What logic do u use to assume them equal? Thanks.
- MathsIsFun
- 2005-09-08 20:21:52
(Atled - great post again - tell us more about yourself!)
- Atled
- 2005-09-08 20:12:33
A . (A + 3B +7C) = A.A + 3(A.B) + 7(A.C)
A.B and A.C are given
A . (A + 3B +7C) = A.A +(3*3) +(7*-1)
A . (A + 3B +7C) = A.A + 2
whats A.A
X.Y = |X|*|Y| * cos(θ)
A is PARALLEL to A there for θ = 0
or
The angle between A and A is 0
cos(0) = 1
A.A = 4
A . (A + 3B +7C) = 4 + 2 =6
hope that helps
edit: previously i said normal
- Mukmin
- 2005-09-08 12:03:24
How do you solve this:
If a.b = 3, a.c = -1, and |a| = 2
what is a.(a+3b+7c)? I couldn't find "a" no matter what way i try it. The only way i could think of is
by setting a/|a| = 1