Dot products?

  • #1
Vectors A, B and C form a triangle. The angle between A and B is [tex]\theta[/tex] and the vectors are related by C=A-B. Compute C . C in terms of the magnitudes A, B and the angle [tex]\theta[/tex] and derive the law of cosines [tex]C^2 = A^2 + B^2 - 2AB cos \theta [/tex].

I have no idea what C . C means but I asked and the lecturer said it was the dot product or something along those lines. Any help would be appreciated :blushing:
 
Last edited:

Answers and Replies

  • #2
xanthym
Science Advisor
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jungleismassiv said:
Vectors A, B and C form a triangle. The angle between A and B is [tex]\theta[/tex] and the vectors are related by C=A-B. Compute C . C in terms of the magnitudes A, B and the angle [tex]\theta[/tex] and derive the law of cosines [tex]C^2 = A^2 + B^2 - 2AB cos \theta [/tex].

I have no idea what C . C means but I asked and the lecturer said it was the dot product or something along those lines. Any help would be appreciated :blushing:
(Note: You should review the properties of vectors and dot products which are discussed in your textbook.)
The dot product (symbol "⋅") between two vectors A and B is defined to be the scalar quantity given by the following equation:
AB = A*B*cos(θ)
where "A" and "B" are the (positive) magnitudes of the respective vectors and where "θ" is the angle between the 2 vectors placed with their initial points together.

Let us compute the dot product of vector C (defined in the problem statement) with itself. Since the angle between any vector (like C) and itself is (0 deg), and since dot product is both distributive and commutative, we can write:
CC = C*C*cos(0) =
= C2 =
= (A - B)⋅(A - B) = ::: ←(From problem statement)
= (AA) - (AB) - (BA) + (BB) =
= A*A*cos(0) - 2*A*B*cos(θ) + B*B*cos(0)
where θ is the angle between vectors A and B.

Simplifying the above equation, we get the required result:
C2 = A2 + B2 - 2*A*B*cos(θ)


~~
 
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  • #3
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