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## #1 2005-09-07 14:03:24

Mukmin
Guest

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

## #2 2005-09-07 22:12:33

Atled
Member
Registered: 2005-08-22
Posts: 9

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

Last edited by Atled (2005-09-09 12:52:58)

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## #3 2005-09-07 22:21:52

MathsIsFun
Registered: 2005-01-21
Posts: 7,684

(Atled - great post again - tell us more about yourself!)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #4 2005-09-08 00:59:49

almukmin
Member
Registered: 2005-09-07
Posts: 2

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.

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## #5 2005-09-08 09:17:22

MathsIsFun
Registered: 2005-01-21
Posts: 7,684

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

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #6 2005-09-08 10:19:40

almukmin
Member
Registered: 2005-09-07
Posts: 2

I just got an answer from my teacher. a.a is basically |a|^2

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## #7 2005-09-09 12:54:40

Atled
Member
Registered: 2005-08-22
Posts: 9