Section 2.9 Dot Product

Engineering Mechanics: Statics Fifteenth Edition Chapter 2 Force Vectors Section 2.9 Dot Product Copyright © 2022 Pearson Education, Inc. All Rights Reserved Dot Product Today’s Objectives: a. determine an angle between two vectors and, b. determine the projection of a vector along a specified line. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Definition (1 of 2) The dot product of vectors A and B is defined as A • B = A B cos . The angle  is the smallest angle between the two vectors and is always in a range of 0 degree to 180 degree. Dot Product Characteristics: 1. The result of the dot product is a scalar (a positive or negative number). 2. The units of the dot product will be the product of the units of the A and B vectors. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Definition (2 of 2) Examples: By definition, i.J = 0 i.i = 1 A g B = (A x i + A y j + A z k ) g (B x i + B y j + B z k ) = A xB x + A yB y + A zB z Copyright © 2022 Pearson Education, Inc. All Rights Reserved Using the Dot Product to Determine the Angle Between Two Vectors For these two vectors in Cartesian form, one can find the angle by a. Find the dot product, A gB = ( A xB x + A yB y + A zBz ) , b. Find the magnitudes (A & B) of the vectors A & B, and c. Use the definition of dot product and solve for , i.e., é(AgB)ù ú, where 0° £  £ 180°.  = cos êê ú êë(A B) úû - 1 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Determining the Projection of a Vector (1 of 2) You can determine the components of a vector parallel and perpendicular to a line using the dot product. Steps: 1. Find the unit vector, ua along line a a 2. Find the scalar projection of A along line aa by AP = A g u a = A x u x + A y u y + A zu z Copyright © 2022 Pearson Education, Inc. All Rights Reserved Determining the Projection of a Vector (2 of 2) ∥ 3. If needed, the projection can be written as a vector, A , by using the unit vector ua and the magnitude found in step 2. AP = AP ua 4. The scalar and vector forms of the perpendicular component can easily be obtained by A ^ = (A - A 2 1 2 2 P ) and ∥ A⊥ = A - A ∥ (rearranging the vector sum of A = A⊥ + A ) Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ (1 of 3) Given: The force acting on the hook at point A. Find: The angle between the force vector and the line A O, and the magnitude of the projection of the force along the line A O. Plan: 1. Find rAO ìï (F g r )ü ï AO ï ï ý 2. Find the angle  = cos íï F r ïï îï ( AO ) þ - 1 3. Find the projection via FAO = F g u AO (or F cos ) Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ (2 of 3) rAO = −1i + 2 j − 2k  m rAO = ( −1) 2 + 22 + ( −2 )  1 2 2 =3m F = −6i + 9 j + 3k  kN F= ( −6) 2 + 92 + 3  1 2 2 = 11.22 kN F g rAO = (- 6)(- 1) + (9)(2) + (3)(- 2) = 18 kN ×m  ( F • rAO )   = cos   F r ( )   AO   18 −1   = cos   = 57.67°  (11.22  3 )  −1 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ (3 of 3) u AO rAO  1  2  2 = = − i +   j + −  K rAO  3  3  3  1 2  2 − 6 − + 9 + 3 ( )   ( )   ( )  −  = 6.00 kN  3 3  3 = F cos  =11.22 cos (57.67) = 6.00 kN FAO = F • u AO = Or : FAO Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (1 of 3) Given: The force acting on the pole at point A. Find: The components of the force acting parallel and perpendicular to the axis of the pole. Plan: 1. Find F , rOA and uOA 2. Determine the parallel component of F using FP = F g uOA 1 3. The perpendicular component of F is F = (F2 - F 2 ) 2 ^ P Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (2 of 3) rOA = −4i + 4 j + 2k  m rOA = ( −4) 2 + 42 + 2  1 2 2 =6m 2 2 1 uOA = − i + j + k 3 3 3 − ( 600 cos 60 ) sin30i  F = + ( 600 cos 60 ) cos30 j  + ( 600 sin 60 ) k    lb   F = −150i + 259.8 j + 519.6K  lb The parallel component of F : ìï 2 2 1 ü ï FP = F • uOA = {- 150i + 259.8 j + 519.6k } • í - i + j + k ý ïïî 3 3 3 ïïþ æ 2 ö÷ æ2 ö÷ æ1 ö÷ ç ç FP = (- 150) ´ ç- ÷ + 259.8 ´ ç ÷ + 519.6 ´ çç ÷ = 446 lb ÷ çè 3 ÷ ç çè3 ÷ ø è3 ø ø Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ (3 of 3) F = 600 lb FP = 446 lb The perpendicular component of F: 1 2 2 P F⊥ = (F2 - F ) = (6002 - 446 1 2 2 ) = 401 lb Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example III Copyright © 2022 Pearson Education, Inc. All Rights Reserved Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (1 of 2) 1. If a dot product of two non-zero vectors is 0, then the two vectors must be ____________ to each other. A. Parallel (pointing in the same direction) B. Parallel (pointing in the opposite direction) C. Perpendicular D. Cannot be determined. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Concept Quiz (2 of 2) 2. If a dot product of two non-zero vectors equals -1, then the vectors must be __________ to each other. A. Collinear but pointing in the opposite direction B. Parallel (pointing in the opposite direction) C. Perpendicular D. Cannot be determined. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 1. The dot product can be used to find all of the following except _______. A. sum of two vectors B. angle between two vectors C. component of a vector parallel to another line D. component of a vector perpendicular to another line Copyright © 2022 Pearson Education, Inc. All Rights Reserved Attention Quiz (2 of 2) 2. Find the dot product of the two vectors P and Q. P = 5i + 2 j + 3k  m Q = −2i + 5 j + 4k  m A. −12 m B. 12 m C. 12 m2 D. −12 m2 E. 10 m2 Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. The dot product of two vectors P and Q is defined as A. P Q sin  B. P Q cos  C. P Q tan  D. P Q sec  Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 2. The dot product of two vectors results in a __________ quantity. A. Scalar B. Vector C. Complex D. Zero Copyright © 2022 Pearson Education, Inc. All Rights Reserved