82.1 Dot Product Concept Overview
DOT PRODUCT | CONCEPT OVERVIEW The topic of the DOT PRODUCT can be referenced on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: The DOT PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the CROSS PRODUCT. The GENERAL FORMULA for the DOT PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors 𝐴 =< 𝑎! , 𝑎! , 𝑎! > and 𝐵 =< 𝑏! , 𝑏! , 𝑏! > with an angle 𝜃 between them, where 𝜃 is between 0 and 𝜋, the DOT PRODUCT can be determined using the following formula: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏! = 𝐴 𝐵 cos 𝜃 = 𝐵 ∙ 𝐴 The difference between the DOT PRODUCT and the CROSS PRODUCT is that the result of a DOT PRODUCT is a SINGLE SCALAR VALUE, which is why it is sometimes referred to as the SCALAR PRODUCT. When given two VECTORS and the ANGLE between those VECTORS, we are given a formula to determine what the DOT PRODUCT will be, however, as you can see, we
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will need to know how to determine the MAGNITUDE of each VECTOR independently. The MAGNITUDE of a VECTOR, can be determined using the formula:
𝐴 =
𝑎! ! + 𝑎! ! + 𝑎! !
The DOT PRODUCT proves useful in physics calculations when we quickly need to determine the action of one vector along the line of action of another vector, this is referred to as the VECTOR PROJECTION. A VECTOR PROJECTION quantifies the portion of one VECTOR that will act in the same DIRECTION of a second vector. This PROJECTION is a new vector that is SCALED DOWN along the same LINE OF ACTION as the VECTOR being projected on. The ORTHOGONAL PROJECTION of any VECTOR U on to any VECTOR V can be graphically illustrated as:
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The ORTHOGONAL PROJECTION of VECTOR U on to VECTOR V is obtained by moving one end of VECTOR U onto VECTOR V and dropping a line from the tip of VECTOR U down to the LINE OF ACTION of VECTOR V. The resulting segment from the TAIL of VECTOR U, along the LINE OF ACTION of VECTOR V and up to the INTERSECTING line that was dropped perpendicularly down to VECTOR V, is referred to as the PROJECTION of VECTOR U on to VECTOR V. In formulaic terms, given VECTOR U and VECTOR V, we can determine the PROJECTION of VECTOR U on to VECTOR V using:
𝑃𝑟𝑜𝑗! 𝑣 =
𝑢∙𝑣 𝑢 𝑢
It is important to note that the reverse of this PROJECTION, where we project VECTOR V on to VECTOR U, will not be same, and represented in formulaic terms as:
𝑃𝑟𝑜𝑗! 𝑢 =
𝑢∙𝑣 𝑣 𝑣
The FORMULA for the COMMUTATIVE PROPERTY of the DOT PRODUCT can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The COMMUTATIVE PROPERTY of the DOT PRODUCT states that the ORDER DOES NOT MATTER when calculating the numerical value of the DOT PRODUCT: 𝐴∙𝐵 =𝐵∙𝐴
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𝐴∙ 𝐵+𝐶 =𝐴∙𝐵+𝐴∙𝐶 The DISTRIBUTIVE PROPERTY of VECTOR ADDITION is represented by the expression: 𝑎 𝐴 + 𝐵 = 𝑎𝐴 + 𝑎𝐵 Where 𝑎 is a scalar value. The FORMULAS for a DOT PRODUCT of a VECTOR and ITSELF can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. As the angle between a vector and itself is zero, and the cosine of zero is one, the magnitude of a vector can be written in terms of the DOT PRODUCT as: 𝐴∙𝐴 = 𝐴
!
𝑖∙𝑖 =𝑗∙𝑗 =𝑘∙𝑘 =1 The FORMULAS for the DOT PRODUCT of TWO ORTHOGONAL VECTORS can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Since the cosine of 90° is zero, the DOT PRODUCT of two orthogonal vectors will result in zero.
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Otherwise simply expressed through the relationship: 𝑖∙𝑗 =𝑗∙𝑘 =𝑘∙𝑖 =0 If 𝐴 ∙ 𝐵 = 0, then either 𝐴 = 0, 𝐵 = 0, or 𝐴 is perpendicular to 𝐵.
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DOT PRODUCT | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Given 𝐴 = 𝑖 and 𝐵 = 2𝑖 + 2𝑗, the dot product, 𝐴 ∙ 𝐵, is most close to: A. 9 B. 8 C. 2 D. 11
SOLUTION: The DOT PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the CROSS PRODUCT. The GENERAL FORMULA for the DOT PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors 𝐴 =< 𝑎! , 𝑎! , 𝑎! > and 𝐵 =< 𝑏! , 𝑏! , 𝑏! > with an angle 𝜃 between them, where 𝜃 is between 0 and 𝜋, the DOT PRODUCT can be determined using the following formula: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏! = 𝐴 𝐵 cos 𝜃 = 𝐵 ∙ 𝐴
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The difference between the DOT PRODUCT and the CROSS PRODUCT is that the result of a DOT PRODUCT is a SINGLE SCALAR VALUE, which is why it is sometimes referred to as the SCALAR PRODUCT. In this problem, we are given: 𝐴=𝑖 𝐵 = 2𝑖 + 2𝑗 We can rewrite these VECTORS as: 𝐴 = 𝑖 + 0𝑗 + 0𝑘 𝐵 = 2𝑖 + 2𝑗 + 0𝑘 Or: 𝐴 =< 1,0,0 > 𝐵 =< 2,2,0 > These VECTORS are now in a form that fit nicely with our GENERAL FORMULA. Revisiting this formula, the DOT PRODUCT can be found using: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏!
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And plugging in our data we get: 𝐴∙𝐵 =1 2 +0 2 +0 0 Or: 𝐴∙𝐵 =2 The correct answer choice is C.
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CONCEPT INTRO: The DOT PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the CROSS PRODUCT. The GENERAL FORMULA for the DOT PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors 𝐴 =< 𝑎! , 𝑎! , 𝑎! > and 𝐵 =< 𝑏! , 𝑏! , 𝑏! > with an angle 𝜃 between them, where 𝜃 is between 0 and 𝜋, the DOT PRODUCT can be determined using the following formula: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏! = 𝐴 𝐵 cos 𝜃 = 𝐵 ∙ 𝐴 The difference between the DOT PRODUCT and the CROSS PRODUCT is that the result of a DOT PRODUCT is a SINGLE SCALAR VALUE, which is why it is sometimes referred to as the SCALAR PRODUCT. When given two VECTORS and the ANGLE between those VECTORS, we are given a formula to determine what the DOT PRODUCT will be, however, as you can see, we
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will need to know how to determine the MAGNITUDE of each VECTOR independently. The MAGNITUDE of a VECTOR, can be determined using the formula:
𝐴 =
𝑎! ! + 𝑎! ! + 𝑎! !
The DOT PRODUCT proves useful in physics calculations when we quickly need to determine the action of one vector along the line of action of another vector, this is referred to as the VECTOR PROJECTION. A VECTOR PROJECTION quantifies the portion of one VECTOR that will act in the same DIRECTION of a second vector. This PROJECTION is a new vector that is SCALED DOWN along the same LINE OF ACTION as the VECTOR being projected on. The ORTHOGONAL PROJECTION of any VECTOR U on to any VECTOR V can be graphically illustrated as:
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The ORTHOGONAL PROJECTION of VECTOR U on to VECTOR V is obtained by moving one end of VECTOR U onto VECTOR V and dropping a line from the tip of VECTOR U down to the LINE OF ACTION of VECTOR V. The resulting segment from the TAIL of VECTOR U, along the LINE OF ACTION of VECTOR V and up to the INTERSECTING line that was dropped perpendicularly down to VECTOR V, is referred to as the PROJECTION of VECTOR U on to VECTOR V. In formulaic terms, given VECTOR U and VECTOR V, we can determine the PROJECTION of VECTOR U on to VECTOR V using:
𝑃𝑟𝑜𝑗! 𝑣 =
𝑢∙𝑣 𝑢 𝑢
It is important to note that the reverse of this PROJECTION, where we project VECTOR V on to VECTOR U, will not be same, and represented in formulaic terms as:
𝑃𝑟𝑜𝑗! 𝑢 =
𝑢∙𝑣 𝑣 𝑣
The FORMULA for the COMMUTATIVE PROPERTY of the DOT PRODUCT can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The COMMUTATIVE PROPERTY of the DOT PRODUCT states that the ORDER DOES NOT MATTER when calculating the numerical value of the DOT PRODUCT: 𝐴∙𝐵 =𝐵∙𝐴
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𝐴∙ 𝐵+𝐶 =𝐴∙𝐵+𝐴∙𝐶 The DISTRIBUTIVE PROPERTY of VECTOR ADDITION is represented by the expression: 𝑎 𝐴 + 𝐵 = 𝑎𝐴 + 𝑎𝐵 Where 𝑎 is a scalar value. The FORMULAS for a DOT PRODUCT of a VECTOR and ITSELF can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. As the angle between a vector and itself is zero, and the cosine of zero is one, the magnitude of a vector can be written in terms of the DOT PRODUCT as: 𝐴∙𝐴 = 𝐴
!
𝑖∙𝑖 =𝑗∙𝑗 =𝑘∙𝑘 =1 The FORMULAS for the DOT PRODUCT of TWO ORTHOGONAL VECTORS can be referenced under the topic of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Since the cosine of 90° is zero, the DOT PRODUCT of two orthogonal vectors will result in zero.
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Otherwise simply expressed through the relationship: 𝑖∙𝑗 =𝑗∙𝑘 =𝑘∙𝑖 =0 If 𝐴 ∙ 𝐵 = 0, then either 𝐴 = 0, 𝐵 = 0, or 𝐴 is perpendicular to 𝐵.
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DOT PRODUCT | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Given 𝐴 = 𝑖 and 𝐵 = 2𝑖 + 2𝑗, the dot product, 𝐴 ∙ 𝐵, is most close to: A. 9 B. 8 C. 2 D. 11
SOLUTION: The DOT PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the CROSS PRODUCT. The GENERAL FORMULA for the DOT PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given two vectors 𝐴 =< 𝑎! , 𝑎! , 𝑎! > and 𝐵 =< 𝑏! , 𝑏! , 𝑏! > with an angle 𝜃 between them, where 𝜃 is between 0 and 𝜋, the DOT PRODUCT can be determined using the following formula: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏! = 𝐴 𝐵 cos 𝜃 = 𝐵 ∙ 𝐴
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The difference between the DOT PRODUCT and the CROSS PRODUCT is that the result of a DOT PRODUCT is a SINGLE SCALAR VALUE, which is why it is sometimes referred to as the SCALAR PRODUCT. In this problem, we are given: 𝐴=𝑖 𝐵 = 2𝑖 + 2𝑗 We can rewrite these VECTORS as: 𝐴 = 𝑖 + 0𝑗 + 0𝑘 𝐵 = 2𝑖 + 2𝑗 + 0𝑘 Or: 𝐴 =< 1,0,0 > 𝐵 =< 2,2,0 > These VECTORS are now in a form that fit nicely with our GENERAL FORMULA. Revisiting this formula, the DOT PRODUCT can be found using: 𝐴 ∙ 𝐵 = 𝑎! 𝑏! + 𝑎! 𝑏! + 𝑎! 𝑏!
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And plugging in our data we get: 𝐴∙𝐵 =1 2 +0 2 +0 0 Or: 𝐴∙𝐵 =2 The correct answer choice is C.
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