Physics 133 Electromagnetism Vectors Analysis MARLON FLORES SACEDON

What the resultant in 3-Dimensions? Vector analysis Recall the component method: 𝑅 𝑥 =− 𝐹 1𝑥 + 𝐹 2𝑥 +… 𝐹 1 𝜃 𝛽 𝐹 2 𝑦 𝑥 𝑅 𝑦 = 𝐹 1𝑦 + 𝐹 2𝑦 +… 𝑅= Σ𝑅 𝑥 2 + Σ𝑅 𝑦 2 Magnitude of resultant 𝛾= 𝑇𝑎𝑛 −1 𝑅 𝑦 𝑅 𝑥 direction of resultant These formulas are for 2-Dimensions only! What the resultant in 3-Dimensions? 𝑧

Vector analysis In 3-Dimensions 𝑅 𝑥 =− 𝐹 1𝑥 + 𝐹 2𝑥 +… 𝑧 𝑥 𝑦 𝑅 𝑥 =− 𝐹 1𝑥 + 𝐹 2𝑥 +… 𝑅 𝑦 = 𝐹 1𝑦 + 𝐹 2𝑦 +… 𝑅 𝑧 = 𝐹 1𝑧 + 𝐹 2𝑧 +… 𝐹 1 Magnitude of resultant 𝑅= Σ𝑅 𝑥 2 + Σ𝑅 𝑦 2 + Σ𝑅 𝑧 2 𝐹 2 𝛾= 𝑇𝑎𝑛 −1 𝐴 𝑦 𝐴 𝑥 directions of resultant 𝜑= 𝑇𝑎𝑛 −1 𝐴 𝑧 𝐴 𝑥 𝛼= 𝑇𝑎𝑛 −1 𝐴 𝑧 𝐴 𝑦

Vector analysis Vectors in term of UNIT VECTORS In 3-dimensions What is Unit Vector? 𝑧 𝑥 𝑦 A unit vector is a vector that has a magnitude of 1, with no units. Its only purpose is to point—that is, to describe a direction in space. A unit vector is often denoted by a lowercase letter with a circumflex, or caret, or "hat": (∧). 𝑗 𝑢 : Unit vector of vector 𝑢 𝑢 : Vector 𝑢 𝑘 𝑖 𝑢 : Magnitude of vector 𝑢 Normalized vectors 𝑢 = 𝑢 𝑢

Vector analysis Vectors in term of UNIT VECTORS 𝑟 𝑟 + 𝑟 𝑦 𝑗 𝑖 + 𝑟 𝑥 + 𝑟 𝑦 𝑟 𝑦 𝑟 𝑟 𝑟 𝑦 = 𝑟 𝑦 𝑗 𝑗 𝜃 𝜃 𝑥 𝑖 + 𝑟 𝑥 𝑟 𝑥 𝑟 𝑥 = 𝑟 𝑥 𝑖 𝑟 = 𝑟 𝑥 + 𝑟 𝑦 𝑟 = 𝑟 𝑥 𝑖 + 𝑟 𝑦 𝑗

Vector analysis Vectors in term of UNIT VECTORS In 3-dimensions 𝑟 𝑗 𝑖 𝑧 𝑥 𝑦 𝑟 = 𝑟 𝑥 + 𝑟 𝑦 + 𝑟 𝑧 𝑟 𝑧 𝑘 𝑟 𝑦 𝑗 𝑟 = 𝑟 𝑥 𝑖 + 𝑟 𝑦 𝑗 + 𝑟 𝑧 𝑘 𝑟 𝑗 𝑟 𝑥 𝑖 𝑘 𝑖

Vector analysis Vector Sum in terms of unit vector Suppose: 𝐴 + 𝐵 = Find: Addition of vectors: 𝐴 + 𝐵 Subtraction of vectors: 𝐴 − 𝐵 𝐴 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 𝐵 = 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 Addition of vectors: 𝐴 + 𝐵 𝐴 + 𝐵 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 + 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 = 𝐴 𝑥 + 𝐵 𝑥 𝑖 + 𝐴 𝑦 + 𝐵 𝑦 𝑗 + 𝐴 𝑧 + 𝐵 𝑧 𝑘 (b) Subtraction of vectors: 𝐴 − 𝐵 𝐴 − 𝐵 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 − 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 = 𝐴 𝑥 − 𝐵 𝑥 𝑖 + 𝐴 𝑦 − 𝐵 𝑦 𝑗 + 𝐴 𝑧 − 𝐵 𝑧 𝑘

Vector analysis Vector Sum in terms of unit vector Example: 5 𝑖 −3 𝑗 + −6 𝑖 + 𝑗 − 1 2 𝑘 Example: = 5−6 𝑖 + −3+1 𝑗 + 0− 1 2 𝑘 𝐶 =5 𝑖 −3 𝑗 𝐷 =−6 𝑖 + 𝑗 − 1 2 𝑘 =− 𝑖 −2 𝑗 − 1 2 𝑘 =− 𝑖 +2 𝑗 + 1 2 𝑘 ANSWER Find: 𝐶 + 𝐷 (b) 𝐶 − 𝐷 Normalized vectors 𝐶 & 𝐷 (b) 𝐶 − 𝐷 = 5 𝑖 −3 𝑗 − −6 𝑖 + 𝑗 − 1 2 𝑘 =5 𝑖 −3 𝑗 +6 𝑖 − 𝑗 + 1 2 𝑘 =11 𝑖 −4 𝑗 + 1 2 𝑘 Note: Normalizing is the same as finding the equivalent unit vector of a given vector. ANSWER (c) 𝐶 = 𝐶 𝐶 = 5 𝑖 −3 𝑗 5 2 + −3 2 =0.86 𝑖 −0.51 𝑗 𝐷 = 𝐷 𝐷 = −6 𝑖 + 𝑗 − 1 2 𝑘 −6 2 + 1 2 + − 1 2 2 =−0.98 𝑖 +0.16 𝑗 −0.08 𝑘 =− 0.98 𝑖 −0.16 𝑗 +0.08 𝑘

Vector analysis Product of vectors Scalar product or Dot product Vector product or Cross product Ex. 𝐴 dot 𝐵 is equal to 𝐴 ∙ 𝐵 Ex. 𝐴 cross 𝐵 is equal to 𝐴 × 𝐵

Vector analysis Product of vectors: The scalar product or Dot product 𝐴 ∙ 𝐵 to be the magnitude of 𝐴 multiplied by the component of 𝐵 in the direction of 𝐴 . 𝐵 ∙ 𝐴 to be the magnitude of 𝐵 multiplied by the component of 𝐴 in the direction of 𝐵 . 𝐵𝑐𝑜𝑠𝜃 𝐴 𝐵 𝜃 𝜃 𝐴 𝐵 𝐴𝑐𝑜𝑠𝜃 𝐵 ∙ 𝐴 =𝐵𝐴𝑐𝑜𝑠𝜃 𝐴 ∙ 𝐵 =𝐴𝐵𝑐𝑜𝑠𝜃 𝐴 ∙ 𝐵 = 𝐴 𝐵 𝑐𝑜𝑠𝜃 𝐵 ∙ 𝐴 = 𝐵 𝐴 𝑐𝑜𝑠𝜃 So, therefore: 𝐴 ∙ 𝐵 = 𝐵 ∙ 𝐴

𝐴 ∙ 𝐵 =𝐴𝐵𝑐𝑜𝑠𝜃 Vector analysis Product of vectors: The scalar product or Dot product So, the definition of Scalar or Dot product is, 𝐴 ∙ 𝐵 =𝐴𝐵𝑐𝑜𝑠𝜃 The scalar product of unit vectors 𝑖 ∙ 𝑖 = 𝑖 𝑖 𝑐𝑜𝑠𝜃 =(1)(1) 𝑐𝑜𝑠 0 𝑜 =1 𝑗 ∙ 𝑗 =1 𝑘 ∙ 𝑘 =1 𝑖 ∙ 𝑗 =0 𝑗 ∙ 𝑖 =0 𝑗 ∙ 𝑘 =0 𝑖 ∙ 𝑘 =0 𝑘 ∙ 𝑖 =0 𝑘 ∙ 𝑗 =0

Vector analysis Product of vectors: The scalar product or Dot product Suppose: Find: 𝐴 ∙ 𝐵 𝐴 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 𝐵 = 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 Then: 𝐴 ∙ 𝐵 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 ∙ 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 1 1 = 𝐴 𝑥 𝐵 𝑥 𝑖 ∙ 𝑖 + 𝐴 𝑥 𝐵 𝑦 𝑖 ∙ 𝑗 + 𝐴 𝑥 𝐵 𝑧 𝑖 ∙ 𝑘 + 𝐴 𝑦 𝐵 𝑥 𝑗 ∙ 𝑖 + 𝐴 𝑦 𝐵 𝑦 𝑗 ∙ 𝑗 + 𝐴 𝑦 𝐵 𝑧 𝑗 ∙ 𝑘 1 + 𝐴 𝑧 𝐵 𝑥 𝑘 ∙ 𝑖 + 𝐴 𝑧 𝐵 𝑦 𝑘 ∙ 𝑗 + 𝐴 𝑧 𝐵 𝑧 𝑘 ∙ 𝑘 𝐴 ∙ 𝐵 = 𝐴 𝑥 𝐵 𝑥 + 𝐴 𝑦 𝐵 𝑦 + 𝐴 𝑧 𝐵 𝑧 Therefore: So: ABcosθ= 𝐴 𝑥 𝐵 𝑥 + 𝐴 𝑦 𝐵 𝑦 + 𝐴 𝑧 𝐵 𝑧 𝜃= 𝑐𝑜𝑠 −1 𝐴 𝑥 𝐵 𝑥 + 𝐴 𝑦 𝐵 𝑦 + 𝐴 𝑧 𝐵 𝑧 𝐴𝐵 Angle between vectors

Vector analysis Product of vectors: The scalar product or Dot product 𝑧 𝑥 𝑦 Example: Find: Given: 𝐺 =9 𝑖 −3 𝑗 + 𝑘 a) 𝐺 ∙ 𝐻 𝐻 𝐻 = 𝑖 −6 𝑘 b) Angle between vectors 𝐺 and 𝐻 −6 𝑘 𝜃 Solution: 𝑖 9 𝑖 𝐺 −3 𝑗 a) 𝐺 ∙ 𝐻 = 𝐺 𝑥 𝐻 𝑥 + 𝐺 𝑦 𝐻 𝑦 + 𝐺 𝑧 𝐻 𝑧 𝑘 = 9 1 + −3 0 +(1)(−6) =3 ANSWER 𝑐𝑜𝑠 −1 𝐺 𝑥 𝐻 𝑥 + 𝐺 𝑦 𝐻 𝑦 + 𝐺 𝑧 𝐻 𝑧 𝐺𝐻 b) 𝜃= =𝑐𝑜𝑠 −1 3 9 2 + −3 2 + 1 2 1 2 + −6 2 =𝑐𝑜𝑠 −1 [ 3 9.49 6.08 ] 𝜃= 87.02 𝑜 ANSWER

𝐴 𝑥 𝐵 = 𝐴 𝐵 𝑠𝑖𝑛𝜃 𝑛 Vector analysis Product of vectors: The vector product or Cross product 𝑛 𝑛 𝐵 𝐴 𝜃 𝐵 𝐴 𝜃 𝐵 𝑠𝑖𝑛𝜃 𝐵 𝑥 𝐴 What is 𝐴 cross 𝐵 is equal to 𝐴 × 𝐵 ? ANSWER: 𝐴 × 𝐵 to be the magnitude of 𝐴 multiplied by the component of 𝐵 perpendicular to 𝐴 , then multiplied by a unit vector 𝑛 which is normal to swipe area of the two vectors. 𝐵 𝑥 𝐴 =− 𝐴 𝑥 𝐵 The property of vector product. 𝐴 𝑥 𝐵 = 𝐴 𝐵 𝑠𝑖𝑛𝜃 𝑛

Vector analysis Product of vectors: The vector product or Cross product The vector product of unit vectors 𝑖 𝑥 𝑖 =1 1 𝑠𝑖𝑛0 𝑛 =0 𝑖 𝑥 𝑗 =1 1 𝑠𝑖𝑛90 𝑘 = 𝑘 𝑗 𝑥 𝑖 =1 1 𝑠𝑖𝑛90 𝑘 =− 𝑘 𝑘 𝑥 𝑘 =0 𝑖 𝑥 𝑖 =0 𝑗 𝑥 𝑗 =0 𝑖 𝑥 𝑗 = 𝑘 𝑗 𝑥 𝑖 =− 𝑘 𝑘 𝑥 𝑗 =− 𝑖 𝑖 𝑥 𝑘 =− 𝑗 𝑗 𝑥 𝑘 = 𝑖 𝑘 𝑥 𝑖 = 𝑗

Vector analysis Product of vectors: The vector product or Cross product Suppose: Find: 𝐴 𝑥 𝐵 𝐴 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 𝐵 = 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 Evaluate: 𝐴 𝑥 𝐵 = 𝐴 𝑥 𝑖 + 𝐴 𝑦 𝑗 + 𝐴 𝑧 𝑘 𝑥 𝐵 𝑥 𝑖 + 𝐵 𝑦 𝑗 + 𝐵 𝑧 𝑘 𝑘 − 𝑗 = 𝐴 𝑥 𝐵 𝑥 𝑖 𝑥 𝑖 + 𝐴 𝑥 𝐵 𝑦 𝑖 𝑥 𝑗 + 𝐴 𝑥 𝐵 𝑧 𝑖 𝑥 𝑘 − 𝑘 𝑖 + 𝐴 𝑦 𝐵 𝑥 𝑗 𝑥 𝑖 + 𝐴 𝑦 𝐵 𝑦 𝑗 𝑥 𝑗 + 𝐴 𝑦 𝐵 𝑧 𝑗 𝑥 𝑘 + 𝐴 𝑧 𝐵 𝑥 𝑘 𝑥 𝑖 𝑗 + 𝐴 𝑧 𝐵 𝑦 𝑘 𝑥 𝑗 − 𝑖 + 𝐴 𝑧 𝐵 𝑧 𝑘 𝑥 𝑘 = 𝐴 𝑥 𝐵 𝑦 𝑘 + 𝐴 𝑥 𝐵 𝑧 − 𝑗 + 𝐴 𝑦 𝐵 𝑥 − 𝑘 + 𝐴 𝑦 𝐵 𝑧 𝑖 + 𝐴 𝑧 𝐵 𝑥 𝑗 + 𝐴 𝑧 𝐵 𝑦 − 𝑖 𝐴 𝑥 𝐵 = 𝐴 𝑦 𝐵 𝑧 𝑖 + 𝐴 𝑧 𝐵 𝑥 𝑗 + 𝐴 𝑥 𝐵 𝑦 𝑘 − 𝐴 𝑦 𝐵 𝑥 𝑘 − 𝐴 𝑧 𝐵 𝑦 𝑖 − 𝐴 𝑥 𝐵 𝑧 𝑗 𝐴 𝑥 𝐵 = 𝑖 𝑗 𝑘 𝐴 𝑥 𝐴 𝑦 𝐴 𝑧 𝐵 𝑥 𝐵 𝑦 𝐵 𝑧 𝐴 𝑥 𝐵 = 𝑖 𝑗 𝑘 𝐴 𝑥 𝐴 𝑦 𝐴 𝑧 𝐵 𝑥 𝐵 𝑦 𝐵 𝑧 𝑖 𝑗 𝐴 𝑥 𝐴 𝑦 𝐵 𝑥 𝐵 𝑦

Vector analysis Product of vectors: The vector product or Cross product 𝐴 𝑥 𝐵 = 𝐴 𝑦 𝐵 𝑧 𝑖 + 𝐴 𝑧 𝐵 𝑥 𝑗 + 𝐴 𝑥 𝐵 𝑦 𝑘 − 𝐴 𝑦 𝐵 𝑥 𝑘 − 𝐴 𝑧 𝐵 𝑦 𝑖 − 𝐴 𝑥 𝐵 𝑧 𝑗 𝐴 𝑥 𝐵 = 𝑖 𝑗 𝑘 𝐴 𝑥 𝐴 𝑦 𝐴 𝑧 𝐵 𝑥 𝐵 𝑦 𝐵 𝑧

Vector analysis Product of vectors: The vector product or Cross product 𝐴 𝑥 𝐵 = 𝐴 𝑦 𝐵 𝑧 𝑖 + 𝐴 𝑧 𝐵 𝑥 𝑗 + 𝐴 𝑥 𝐵 𝑦 𝑘 − 𝐴 𝑦 𝐵 𝑥 𝑘 − 𝐴 𝑧 𝐵 𝑦 𝑖 − 𝐴 𝑥 𝐵 𝑧 𝑗 𝐴 𝑥 𝐵 = 𝑖 𝑗 𝑘 𝐴 𝑥 𝐴 𝑦 𝐴 𝑧 𝐵 𝑥 𝐵 𝑦 𝐵 𝑧 𝐴 𝑥 𝐵 = 𝐴 𝑦 𝐵 𝑧 𝑖 − 𝐴 𝑧 𝐵 𝑦 𝑖 + 𝐴 𝑧 𝐵 𝑥 𝑗 − 𝐴 𝑥 𝐵 𝑧 𝑗 + 𝐴 𝑥 𝐵 𝑦 𝑘 − 𝐴 𝑦 𝐵 𝑥 𝑘 𝐴 𝑥 𝐵 = 𝐴 𝑦 𝐵 𝑧 − 𝐴 𝑧 𝐵 𝑦 𝑖 + 𝐴 𝑧 𝐵 𝑥 − 𝐴 𝑥 𝐵 𝑧 𝑗 + 𝐴 𝑥 𝐵 𝑦 − 𝐴 𝑦 𝐵 𝑥 𝑘 𝐴 𝑥 𝐵 = 𝑖 𝑗 𝑘 𝐴 𝑥 𝐴 𝑦 𝐴 𝑧 𝐵 𝑥 𝐵 𝑦 𝐵 𝑧

Vector analysis Product of vectors: The vector product or Cross product Example: Given: Find: 𝐹 =3 𝑖 −5 𝑗 +2 𝑘 a) 𝐹 × 𝐺 b) 𝐺 × 𝐹 𝐺 = 𝑗 −7 𝑘 𝐹 𝑥 𝐺 = 𝑖 𝑗 𝑘 3 −5 2 0 1 −7 = 35−2 𝑖 + 0+21 𝑗 + 3−0 𝑘 =33 𝑖 +21 𝑗 +3 𝑘 𝐺 𝑥 𝐹 = 𝑖 𝑗 𝑘 0 1 −7 3 −5 2 = 2−35 𝑖 + −21−0 𝑗 + 0−3 𝑘 =−33 𝑖 −21 𝑗 −3 𝑘 =− 33 𝑖 +21 𝑗 +3 𝑘 So: 𝐺 × 𝐹 =− 𝐹 × 𝐺

Assignment 𝐴 𝐵 𝐶 1) Given: 2) Figure: Find: 300𝑔𝑓 Find: 100𝑔𝑓 160𝑔𝑓 𝐿 = 𝑖 +6 𝑗 −3 𝑘 𝐴 𝐵 𝑀 =6 𝑖 − 𝑗 +7 𝑘 100𝑔𝑓 25 𝑜 160𝑔𝑓 𝑁 =−7 𝑘 65 𝑜 300𝑔𝑓 40 𝑜 Find: a) 𝐿 × 𝑁 b) 𝐿 × 𝑀 c) 𝐿 ∙ 𝑀 × 𝑁 d) 𝐿 × 𝑀 ∙ 𝑀 × 𝑁 e) 𝐿 × 𝑀 × 𝑁 𝐶 Find: f) 5 𝑀 × 𝑁 a) Convert the three vectors in terms of unit vectors. g) 5 𝑁 × 𝑀 b) Find the vector sum ( 𝐴 + 𝐵 + 𝐶 ) Note: Use the converted vectors in terms of unit vectors to find the sum. h) Find the unit vectors of 𝐿 , 𝑁 and 𝑀 . i) Graph the vectors 𝐿 , 𝑁 and 𝑀 .

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