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About myself...25/m/CaEE student at UC Davishobbies: cycling, skating, surfing although there are no oceans near davis

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

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

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 + 2whats 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 0cos(0) = 1A.A = 4A . (A + 3B +7C) = 4 + 2 =6 hope that helps

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 + 2whats 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 0cos(0) = 1A.A = 4A . (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.

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

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 + 2whats 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 0cos(0) = 1A.A = 4A . (A + 3B +7C) = 4 + 2 =6 hope that helpsedit: previously i said normal

How do you solve this:If a.b = 3, a.c = -1, and |a| = 2what 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