1 / 23

Dot Product

This slideshow provides a comprehensive review of the Dot Product of two vectors, including its definition, basic examples, laws of operations, identities, component calculations, applications in mechanics, and finding parallel and perpendicular components to a given direction.

wcamacho
Download Presentation

Dot Product

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Dot Product This slideshow will be a review on the Dot Product of two vectors

  2. Definition The Dot Product of vectors A and B is defined as A· B = |A| |B| cos Θ B A A Θ B

  3. Simple Example of Dot Product A B From the given Vectors: A = 6i + 8j y B = 8i Find A• B x

  4. Solution to Simple Dot Product |A| = sqrt (62 + 82) = 10 -Find |A| -Find |B| |B| = sqrt ( 82) = 8 We must then find the angle between the vectors (Θ) Θ = tan-1 (8/6)= 53.1° y A 8 Θ Θ 6 B x Use the given equation: A· B = |A| |B| cos Θ A· B = (10)(8) cos 53.1 = 48.03

  5. Laws of Operations • Commutative A · B = B · A • Associative with respect to scalar multiplication a (A · B) = (a A) · B = A · (a B) • Distributive with respect to vector addition A· (B + D) = (A· B) + (A · D)

  6. Dot Product Identities i· i = 1 j· j = 1 k· k = 1 Since the angle between two i direction vectors would be 0, the equation would be: i· i = |i| |i| cos 0° = (1)(1)(1) = 1 i· j = 0 i· k = 0 k· j = 0 j· i = 0 k· i = 0 j· k = 0 Since the angle between two different direction vectors would be 90°, the equation would be: i· j = |i| |j| cos 90° = (1)(1)(0) = 0 Since the cosine of 90° = 0, and the cosine of 0° = 1, from A·B = cos Θ, we can derive:

  7. Using the laws from the previous page, we see that this equation can be reduced to: A· B = AxBx + AyBy + AzBz Components of Dot Product Consider the dot product of two 3 dimensional vectors expressed in Cartesian vector form: A· B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk) = AxBx( i · i ) + AxBy( i · j ) + AxBz( i · k ) + AyBx( j · i ) + AyBy( j · j ) + AyBz( j · k ) + AxBx( k · i ) + AzBy( k · j ) + AzBz( k · k ) Therefore, to determine the Dot Product of two vectors, multiply their corresponding x, y, z components and sum their products.

  8. Example of Finding Dot Product Using Cartesian Components B A z Given two vectors: A = 2i + 3j – 4k B = -3i + 2k Find A • B y x

  9. Solution to Component Dot Product Use the formula A· B = AxBx + AyBy + AzBz A = 2i + 3j – 4k Ax = 2; Ay = 3; Az = -4 B = -3i + 2k Bx = -3; By = 0; Bz = 2 A· B = (2)(-3) + (3)(0) + (-4)(2) = -14

  10. (A• B) Θ = cos -1 |A| |B| Applications The Dot product has useful applications in mechanics • The angle formed between two vectors or intersecting lines • The angle between the tails of vectors A and B can be determined by solving A· B = |A| |B| cos Θ, for Θ.

  11. Example Finding Angle Between Vectors B Θ A Find the angle between the two vectors in the previous example z A = 2i + 3j – 4k B = -3i + 2k y x

  12. Solution for Finding Angle (A• B) Θ = cos -1 |A| |B| First, we will find (A• B) using the component form: A = 2i + 3j – 4k B = -3i + 2k -14 = Θ = cos -1 (5.385)(3.606) Use the equation: A· B = AxBx + AyBy + AzBz = (2)(-3) + (3)(0) + (-4)(2) = -14 We will then find the magnitude of A and B: |A| = sqrt (22 + 32 + (-4)2) = 5.385 |B| = sqrt ( (-3)2 + 02 + 22) = 3.606 136.13°

  13. Vector Parallel Components If the direction is specified by the unit vector u ( since u = 1) we can also determine |A|||from the dot product as: |A||| = |A| cos Θ = A • u 2. The components of a vector parallel to a given direction. • The component of vector A parallel (|A|||) to a direction is defined as: • |A||| = |A| cos Θ Since the dot product produces a scalar quantity, this is the magnitude of the vector parallel to the direction. If A|| is positive, the direction is the same as u. If A|| is negative, the direction is opposite that of u.

  14. Example finding Parallel Component to Direction L F Find the magnitude of the force, F, that is parallel to the direction L. z L = (2i +6j + 3k) meters F = (300j) Newtons y x

  15. Solution Finding Parallel Component Use the equation: |A||| = A • u We must first find the unit vector of the given direction L: The unit vector is defined as: u = L / |L| |L| = sqrt (22 + 62 + 32) = 7 uL = (2i + 6j + 3k) / 7 = .2865i + .857j + .429k |F||| = F • uL = (300j Newtons) • (.2865i + .857j + .429k) = (0i) •(.286i) + (300j) •(.857j) + (0k) •(.429k) = 257.1 Newtons

  16. Vector Parallel Components z Find the vector components of the force, F, parallel to the direction L. L F L = (2i +6j + 3k) meters y F = (300j) Newtons x To obtain the vector form of the component parallel to direction L, the magnitude found by F• u is simply multiplied by the unit vector u. F|| = |F| (cos Θ) u = (F • u) u

  17. Solution Vector Parallel Components In the previous example we found that: |F||| = F • u = 257.1 Newtons uL= .2865i + .857j + .429k Then to find the vector components of F parallel to the direction L, use: A|| = (A • u) u F|| = (F • uL) uL = ((300j) • (.2865i + .857j + .429k))(.2865i + .857j + .429k)) = (73.5i + 220j +110k) N (257.1 Newtons) (.2865i + .857j + .429k) =

  18. Finding Vector Perpendicular Components 3. Finding a vector’s components perpendicular to a given direction A vector that is perpendicular, or normal, to a given direction may be found in either scalar or vector form. To find the vector components of the force perpendicular (Fn) to a direction, A|| must first be found. A||(vector form) is subtracted from the given vector A. An = A - A||

  19. Example F Perpendicular to Direction z L F y x Find the vector components of force, F, that is perpendicular to the direction L. L = (2i +6j + 3k) meters F = (300j) Newtons

  20. Solution to Finding F Perpendicular From a previous example, we found that: F|| = (73.5i + 220j +110k) N We know that: F = (300j) Newtons Use: An = A - A|| Fn = (300j)N – (73.5i + 220j + 110k)N Fn = (-73.5i + 80j – 110k)N

  21. Magnitude of F perpendicular to Direction There are two ways to find the portion of a force perpendicular to a direction If you have already found Fn, the magnitude of the vector form can be taken Given: Fn = Fxi + Fyj + Fzk |F|n= sqrt ( Fx2 + Fy2 + Fz2) The other way is to use the magnitudes of the force and the parallel force: |F|n = sqrt (|F|2 - |F|||2)

  22. z L F y x Example Magnitude of F Perpendicular to Direction Find the magnitude of the force, F, that is perpendicular to the direction L. L = (2i +6j + 3k) meters F = (300j) Newtons

  23. Solution finding magnitude of F perpendicular Using magnitude of Fn: Fn = (-73.5i + 80j – 110k)N |F|n= sqrt ( (-73.5)2 +802 +(-110)2) = 154.60 Newtons Using |F| and |F||| : |F| = 300 N This error comes from when we rounded 6/7 in the unit vector of L from .85714 to .857 |F||| = 257.1 N |F|n= sqrt (3002 – 257.12) = 154.59 Newtons

More Related