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Atled
2005-09-10 10:54:40

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