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Dot Product

Dot Product. This slideshow will be a review on the Dot Product of two vectors. Definition. The Dot Product of vectors A and B is defined as A · B = | A | | B | cos Θ. B. A. A. Θ. B. Simple Example of Dot Product. A. B. From the given Vectors:. A = 6 i + 8 j. y. B = 8 i.

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Dot Product

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  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

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