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CE 201 - Statics

CE 201 - Statics. Lecture 6. Contents. Dot Product. A. B. . DOT PRODUCT. Dot product is the method of multiplying two vectors and is used to solve three-dimensional problems If A and B are two vectors then, A . B = AB cos (  ) where 0    180 . Laws of Operation.

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CE 201 - Statics

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  1. CE 201 - Statics Lecture 6

  2. Contents • Dot Product

  3. A B  DOT PRODUCT Dot product is the method of multiplying two vectors and is used to solve three-dimensional problems If A and B are two vectors then, A . B = AB cos () where 0 180

  4. Laws of Operation (1) Commutative Law A . B = B . A (2) Multiplication by a scalar a  ( A . B ) = ( aA ) . B = A . ( aB ) = ( A . B )  a (3) Distribution Law A . ( B + D ) = ( A . B ) + ( A . D )

  5. Cartesian Vector Formulation i . i = 1 j . j = 1 k . k = 1 i . j = 0 i . k = 0 k . j = 0 i . i = (1)  (1) cos 0 = 1 i . k = (1)  (1) cos 90 = 0 If A and B are Cartesian vectors, then A . B = (Ax i + Ay j + Az k) . (Bx i + By j + Bz k = Ax Bx (i.i) + Ax By (i.j) + Ax Bz (i.k) Ay Bx (j.i) + Ay By (j.j) + Ay Bz (j.k) Az Bx (k.i) + Az By (k.j) + Az Bz (k.k) Doing Dot Product, we have A . B = Ax Bx + Ay By + Az Bz[scalar ( so there is no i, j, k)]

  6. Applications (1) to find the angle  formed between two vectors or two intersecting lines. If A and B are vectors, then  between their tails will be:  = cos-1 ( A . B ) / (AB) 0 180 If A . B = 0  = cos-1 (0) = 90 (A is perpendicular to B)

  7. A AI  a a u AII = A cos () u • to find the rectangular components of a vector. AII = A cos () AII = A cos  = A . u AII = A cos () u = (A . u) u AI = A – AII (since A = AII + AI) To find the magnitude of AI If cos () = A . U /A  = cos-1 ( A . u ) / A (0 180) AI = A sin () or AI = A2 – (AII)2 (Pythagorean)

  8. Examples • Examples 2.16 – 2.17 • Problem 2.113 • Problem 2.129

  9. Assignment No. 1 (Chapter 2) 1, 9, 12, 19, 23, 28, 29, 33, 36, 40, 47, 54, 58, 60, 65, 69, 73, 76, 78, 80

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