‫ﺍﻟﻣﺗﺟﻬﺎﺕ‬
‫‪Vectors‬‬
‫ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺻﻝ ﺳﻧﻠﻘﻲ ﻧﻅﺭﺓ ﻣﺧﺗﺻﺭﺓ ﻋﻠﻰ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻭ ﺑﻌﺽ ﺧﻭﺍﺻﻬﺎ‪ .‬ﺳﻧﺗﻧﺎﻭﻝ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺻﻝ‬
‫ﺍﻟﻣﻭﺍﺿﻳﻊ ﺍﻵﺗﻳﺔ‪:‬‬
‫‪. 1‬ﺍﻟﻣﺗﺟﻬﺎﺕ ) ﺑﻌﺽ ﺍﻟﻣﺑﺎﺩﺉ ﺍﻷﻭﻟﻳﺔ (‪.‬‬
‫‪.2‬ﺑﻌﺽ ﺍﻟﻌﻣﻠﻳﺎﺕ ﻋﻠﻰ ﺍﻟﻣﺗﺟﻬﺎﺕ‪.‬‬
‫‪.3‬ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ‪.‬‬
‫‪ .4‬ﺍﻟﺿﺭﺏ ﺃﻻﺗﺟﺎﻫﻲ‪.‬‬
‫ﺍﻟﻣﺗﺟﻬﺎﺕ ) ﺑﻌﺽ ﺍﻟﻣﺑﺎﺩﺉ ﺍﻷﻭﻟﻳﺔ (‬
‫ﺩﻋﻧﺎ ﻧﺑﺩﺃ ﺍﻟﻔﺻﻝ ﺑﻔﺎﺋﺩﺓ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻭ ﻟﻣﺎﺫﺍ ﺗﺳﺗﺧﺩﻡ‪ .‬ﺗﺳﺗﺧﺩﻡ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻟﺗﻣﺛﻳﻝ ﺍﻟﻛﻣﻳﺎﺕ ﺍﻟﺗﻲ ﺗﻣﺗﻠﻙ‬
‫ﻣﻘﺩﺍﺭﺍً ﻭ ﺍﺗﺟﺎﻫﺎ ً‪ .‬ﻛﺎﻟﺳﺭﻋﺔ ﻭ ﺍﻟﻘﻭﺓ ﻣﺛﻼً‪ .‬ﻟﺗﻌﺭﻳﻑ ﺍﻟﻘﻭﺓ ﻧﺣﺗﺎﺝ ﻻﻥ ﻧﻌﺭﻑ ﻣﻘﺩﺍﺭﻫﺎ ﻭ ﺍﺗﺟﺎﻫﻬﺎ‪.‬‬
‫ﻳﻣﻛﻥ ﺃﻥ ﻧﻣﺛﻝ ﺍﻟﻣﺗﺟﻪ ﺑﺧﻁ ﻟﻪ ﺍﺗﺟﺎﻩ ﻳﻣﺛﻝ ﻁﻭﻝ ﺍﻟﺧﻁ ﻣﻘﺩﺍﺭﻩ ﻭ ﺍﺗﺟﺎﻩ ﺍﻟﺧﻁ ﻫﻭ ﺍﺗﺟﺎﻩ ﺍﻟﻣﺗﺟﻪ‪.‬‬
‫ﺃﻥ ﺃﻳﺔ ﻗﻭﺓ ﻣﺳﻠﻁﺔ ﻋﻠﻰ ﻧﻘﻁﺔ ﻣﺎ ﺑﺎﺗﺟﺎﻩ ﻣﻌﻳﻥ ﻳﻣﻛﻥ ﺗﺳﻠﻳﻁﻬﺎ ﻋﻠﻰ ﺃﻳﺔ ﻧﻘﻁﺔ ﻓﻲ ﺍﻟﻔﺿﺎء ﺑﻧﻔﺱ‬
‫ﺍﻻﺗﺟﺎﻩ ﻭ ﺍﻟﻣﻘﺩﺍﺭ‪ .‬ﺃﻱ ﺃﻥ ﺍﻟﻘﻭﺓ ﻻ ﺗﻌﺗﻣﺩ ﻋﻠﻰ ﺍﻟﻧﻘﻁﺔ ﺍﻟﻣﺳﻠﻁﺔ ﻋﻠﻳﻬﺎ‪ .‬ﻻﺣﻅ ﺍﻟﺭﺳﻡ ﻓﻲ ﺍﻷﺳﻔﻝ‪:‬‬
‫‪1‬‬
‫ﺃﻥ ﻛﻝ ﻗﻁﻌﺔ ﻓﻲ ﺍﻟﺭﺳﻡ ﺃﻋﻼﻩ ﺗﻣﺛﻝ ﻧﻔﺱ ﺍﻟﻣﺗﺟﻪ‪ .‬ﻓﻲ ﻛﻝ ﺣﺎﻟﺔ ﻳﺑﺩﺃ ﺍﻟﻣﺗﺟﻪ ﻣﻥ ﻧﻘﻁﺔ ﻣﻌﻳﻧﺔ ﺑﻌﺩﻫﺎ‬
‫ﻳﺗﺣﺭﻙ ﻭﺣﺩﺗﻳﻥ ﺇﻟﻰ ﺍﻟﻳﺳﺎﺭ ﻭ ﺧﻣﺱ ﻭﺣﺩﺍﺕ ﺇﻟﻰ ﺍﻷﻋﻠﻰ‪ .‬ﻭ ﺳﻧﺭﻣﺯ ﻟﻬﺎ ﺑﺎﻟﺭﻣﺯ ⟩‪.𝑣𝑣⃑ = ⟨−2,5‬‬
‫ﻳﻣﺛﻝ ﺍﻟﻣﺗﺟﻪ ﻛﻣﻳﺔ ﻭﺍﺗﺟﺎﻩ ﺍﻟﻣﻘﺩﺍﺭ ﺑﻳﻧﻣﺎ ﺗﻣﺛﻝ ﺍﻟﻧﻘﻁﺔ ﺍﻟﻣﻭﻗﻊ ﻓﻲ ﺍﻟﻣﺳﺗﻭﻱ‪.‬‬
‫ﺃﻥ ﺗﻣﺛﻳﻝ ﺍﻟﻣﺗﺟﻪ ⟩ ‪ 𝑣𝑣⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2‬ﻓﻲ ﺍﻟﻔﺿﺎء ﺫﻱ ﺍﻟﺑﻌﺩﻳﻥ ﻫﻭ ﺍﻳﺔ ﻗﻁﻌﺔ ﻣﺳﺗﻘﻳﻡ ⃑�����‬
‫𝐴𝐴𝐴𝐴 ﻣﻥ ﺍﻟﻧﻘﻁﺔ‬
‫)𝑦𝑦 ‪ 𝐴𝐴 = (𝑥𝑥,‬ﻭ ﺣﺗﻰ ﺍﻟﻧﻘﻁﺔ ) ‪ 𝐵𝐵 = (𝑥𝑥 + 𝑎𝑎1 , 𝑦𝑦 + 𝑎𝑎2‬ﺍﻣﺎ ﺗﻣﺛﻳﻝ ﺍﻟﻣﺗﺟﻪ ⟩ ‪𝑣𝑣⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2, 𝑎𝑎3‬‬
‫ﻓﻲ ﺍﻟﻔﺿﺎء ﺛﻼﺛﻲ ﺍﻻﺑﻌﺎﺩ ﻫﻭ ﻗﻁﻌﺔ ﺍﻟﻣﺳﺗﻘﻳﻡ 𝐴𝐴𝐴𝐴‬
‫⃑����� ﻣﻥ ﺍﻟﻧﻘﻁﺔ )𝑧𝑧 ‪ 𝐴𝐴 = (𝑥𝑥, 𝑦𝑦,‬ﻭ ﺣﺗﻰ ﺍﻟﻧﻘﻁﺔ‬
‫) ‪ .𝐵𝐵 = (𝑥𝑥 + 𝑎𝑎1 , 𝑦𝑦 + 𝑎𝑎2 , 𝑧𝑧 + 𝑎𝑎3‬ﻳﻣﺛﻝ ﺍﻟﻣﺗﺟﻪ‬
‫⟩ ‪ 𝑣𝑣⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﺍﻟﺫﻱ ﻳﺑﺩﺃ ﺑﺎﻟﻧﻘﻁﺔ‬
‫)‪ 𝐴𝐴 = (0,0,0‬ﻭ ﻳﻧﺗﻬﻲ ﺑﺎﻟﻧﻘﻁﺔ ) ‪ . 𝐵𝐵 = (𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﺑﻣﺎ ﻳﺳﻣﻰ ﻣﺗﺟﻪ ﺍﻟﻣﻭﻗﻊ ‪position‬‬
‫‪ vector‬ﻟﻠﻧﻘﻁﺔ ) ‪.(𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬‬
‫ﻭ ﺃﻻﻥ ﻛﻳﻑ ﻧﺳﺗﺧﺭﺝ ﻣﺗﺟﻪ ﻋﻠﻣﺕ ﻧﻘﻁﺗﺎ ﺑﺩﺍﻳﺗﻪ ﻭ ﻧﻬﺎﻳﺗﻪ‪.‬‬
‫ﺃﻥ ﺍﻟﻣﺗﺟﻪ ﺍﻟﺫﻱ ﻳﺑﺩﺃ ﺑـﺎﻟﻧﻘﻁﺔ ) ‪ 𝐴𝐴 = (𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻭ ﻳﻧﺗﻬﻲ ﺑﺎﻟﻧﻘﻁﺔ ) ‪ 𝐵𝐵 = (𝑏𝑏1 , 𝑏𝑏2 , 𝑏𝑏3‬ﻫﻭ‪:‬‬
‫⃑�����‬
‫) ‪𝐴𝐴𝐴𝐴 = 𝑣𝑣⃑ = (𝑏𝑏1 −𝑎𝑎1 , 𝑏𝑏2 − 𝑎𝑎2 , 𝑏𝑏3 − 𝑎𝑎3‬‬
‫ﻻﺣﻅ ﺍﻻﺗﺟﺎﻩ ﺑﺩﻗﺔ ﺇﺫ ﺃﻥ ﺍﻟﻣﺗﺟﻪ ﺍﻟﺫﻱ ﻳﺑﺩﺃ ﺑﺎﻟﻧﻘﻁﺔ ) ‪ 𝐵𝐵 = (𝑏𝑏1 , 𝑏𝑏2 , 𝑏𝑏3‬ﻭ ﻳﻧﺗﻬﻲ ﺑﺎﻟﻧﻘﻁﺔ‬
‫) ‪ 𝐴𝐴 = (𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻳﻣﻛﻥ ﺗﻣﺛﻳﻠﻪ ﺑﺎﻟﺻﻭﺭﺓ‪:‬‬
‫⃑�����‬
‫) ‪𝐵𝐵𝐵𝐵 = 𝑣𝑣⃑ = (𝑎𝑎1 −𝑏𝑏1 , 𝑎𝑎2 − 𝑏𝑏2 , 𝑎𝑎3 − 𝑏𝑏3‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺍﻛﺗﺏ ﺍﻟﻣﺗﺟﻪ ﻟﻛﻝ ﻣﻣﺎ ﻳﺄﺗﻲ‪:‬‬
‫‪ .a‬ﺍﻟﻣﺗﺟﻪ ﻣﻥ )‪ (2, −7,0‬ﺇﻟﻰ )‪. (1, −3, −5‬‬
‫‪ .b‬ﺍﻟﻣﺗﺟﻪ ﻣﻥ )‪ (1, −3, −5‬ﺇﻟﻰ )‪. (2, −7,0‬‬
‫‪ .c‬ﻣﺗﺟﻪ ﺍﻟﻣﻭﻗﻊ ﺇﻟﻰ )‪.(−90,4‬‬
‫ﺍﻟﺣﻝ‪.⟨−1,4, −5⟩ .a .‬‬
‫‪.⟨1, −4,5⟩ .b‬‬
‫ﺃﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ ﻓﻲ ‪ a‬ﻭ ‪ b‬ﻣﺧﺗﻠﻔﻳﻥ ﺑﺎﻹﺷﺎﺭﺓ ﻓﻘﻁ ﻭ ﻫﺫﺍ ﻳﺑﻳﻥ ﺃﻥ ﻟﻬﻣﺎ ﻧﻔﺱ ﺍﻟﻣﻘﺩﺍﺭ ﻟﻛﻧﻬﻣﺎ ﻣﺗﻌﺎﻛﺳﻳﻥ‬
‫ﺑﺎﻻﺗﺟﺎﻩ‪.‬‬
‫‪.⟨−90,4⟩.c‬‬
‫ﺗﻌﺭﻳﻑ‪ .‬ﺃﻥ ﻁﻭﻝ ‪ length‬ﺍﻟﻣﺗﺟﻪ ⟩ ‪ 𝑣𝑣⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻫﻭ‪:‬‬
‫‪1‬‬
‫‪‖𝑣𝑣⃑‖ = (𝑎𝑎1 2 + 𝑎𝑎2 2 + 𝑎𝑎3 2 )2‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺟﺩ ﻁﻭﻝ ﻛﻝ ﻣﻥ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺍﻵﺗﻳﺔ‪:‬‬
‫⟩‪𝑎𝑎. 𝑎𝑎⃑ = ⟨3, −5,10‬‬
‫𝑢𝑢 ‪𝑏𝑏.‬‬
‫⟩‪�⃑ = ⟨1/√5, −2/√5‬‬
‫𝑤𝑤 ‪𝑐𝑐.‬‬
‫⟩‪��⃑ = ⟨0,0‬‬
‫𝑢𝑢 ‪𝑑𝑑.‬‬
‫⟩‪�⃑ = ⟨1,0,0‬‬
‫‪2‬‬
‫‪a.‬ﺍﻟﺣﻝ‪.‬‬
‫‪= √134.‬‬
‫ﺑﺎﻗﻲ ﺍﻟﻔﺭﻭﻉ ﺗﺗﺭﻙ ﻛﺗﻣﺭﻳﻥ ﻟﻠﻁﺎﻟﺏ‪.‬‬
‫‪1‬‬
‫‪100)2‬‬
‫‪‖𝑎𝑎⃑‖ = (9 + 25 +‬‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﺃﺫﺍ ﻛﺎﻥ ‪ ‖𝑎𝑎⃑‖ = 0‬ﻓﺎﻥ ⃑𝑎𝑎 ﻫﻭ ﺍﻟﻣﺗﺟﻪ ﺍﻟﺻﻔﺭﻱ‪.‬‬
‫ﺗﻌﺭﻳﻑ‪ .‬ﺃﻱ ﻣﺗﺟﻪ ﻁﻭﻟﻪ ﻳﺳﺎﻭﻱ ﻭﺍﺣﺩ ﻳﺳﻣﻰ ﻣﺗﺟﻪ ﺍﻟﻭﺣﺩﺓ ‪. unit vector‬‬
‫ﺗﻣﺭﻳﻥ‪ .‬ﺃﻳﺎ ﻣﻥ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻓﻲ ﺍﻟﻣﺛﺎﻝ ﺃﻋﻼﻩ ﻳﻣﺛﻝ ﻣﺗﺟﻪ ﺍﻟﻭﺣﺩﺓ؟‬
‫𝑤𝑤 ﺍﻟﻣﺗﺟﻪ ﺍﻟﺻﻔﺭﻱ ‪.zero vector‬‬
‫ﺗﻌﺭﻳﻑ‪ .‬ﻳﺳﻣﻰ ﺍﻟﻣﺗﺟﻪ ⟩‪��⃑ = ⟨0,0‬‬
‫ﻣﻼﺣﻅﺔ‪.1 .‬ﻓﻲ ﺍﻟﻔﺿﺎءﺍﺕ ﺛﻼﺛﻳﺔ ﺍﻷﺑﻌﺎﺩ ﺗﻭﺟﺩ ﺛﻼﺙ ﻣﺗﺟﻬﺎﺕ ﻟﻠﻘﺎﻋﺩﺓ ﺍﻷﺳﺎﺱ‬
‫‪ basis vector‬ﻭﻫﻲ‪:‬‬
‫⟩‪�⃑ = ⟨0,0,1‬‬
‫𝑘𝑘‬
‫‪standard‬‬
‫‪𝑖𝑖⃑ = ⟨1,0,0⟩ , 𝑗𝑗⃑ = ⟨0,1,0⟩ ,‬‬
‫ﺃﻣﺎ ﻓﻲ ﺍﻟﻔﺿﺎءﺍﺕ ﺛﻧﺎﺋﻳﺔ ﺍﻷﺑﻌﺎﺩ ﻳﻭﺟﺩ ﻣﺗﺟﻬﻳﻥ ﻟﻠﻘﺎﻋﺩﺓ ﺍﻷﺳﺎﺱ ﻫﻣﺎ‪:‬‬
‫⟩‪𝑖𝑖⃑ = ⟨1,0⟩ , 𝑗𝑗⃑ = ⟨0,1‬‬
‫‪ .2‬ﻻ ﺗﻘﺗﺻﺭ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻋﻠﻰ ﺍﻟﻔﺿﺎءﺍﺕ ﺛﻧﺎﺋﻳﺔ ﻭ ﺛﻼﺛﻳﺔ ﺍﻷﺑﻌﺎﺩ ﺑﻝ ﻳﺗﻌﺩﻯ ﺫﻟﻙ ﺇﻟﻰ ﺍﻟﻔﺿﺎءﺍﺕ‬
‫ﺫﺍﺕ ﺍﻟﺑﻌﺩ ‪ n-dimensional space n -‬ﻭ ﺗﻛﻭﻥ ﻣﺗﺟﻬﺎﺗﻬﺎ ﺑﺎﻟﺻﻳﻐﺔ‪:‬‬
‫⟩ 𝑛𝑛𝑎𝑎 ‪𝑣𝑣⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , … ,‬‬
‫ﺃﻥ ﻛﻝ 𝑗𝑗𝑎𝑎 ﺗﺩﻋﻰ ﻣﺭﻛﺑﺔ ‪ component‬ﻟﻠﻣﺗﺟﻪ‪.‬‬
‫ﻓﻲ ﺩﺭﺍﺳﺗﻧﺎ ﺍﻟﺣﺎﻟﻳﺔ ﺳﻧﺗﻌﺎﻣﻝ ﻣﻊ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺛﻧﺎﺋﻳﺔ ﻭ ﺛﻼﺛﻳﺔ ﺍﻟﻣﺭﻛﺑﺎﺕ ﻟﻠﺳﻬﻭﻟﺔ‪.‬‬
‫ﺟﺑﺭ ﺍﻟﻣﺗﺟﻬﺎﺕ ‪vector algebra‬‬
‫ﻧﺑﺩﺃ ﺑﺟﻣﻊ ﺍﻟﻣﺗﺟﻬﺎﺕ‪ :‬ﺃﻥ ﺣﺎﺻﻝ ﺟﻣﻊ ‪ addition‬ﺍﻟﻣﺗﺟﻬﻳﻥ‬
‫ﻳﻌﻁﻰ ﺑﺎﻟﺻﻳﻐﺔ‪:‬‬
‫⟩ ‪ 𝑏𝑏�⃑ = ⟨𝑏𝑏1 , 𝑏𝑏2 , 𝑏𝑏3‬ﻭ ⟩ ‪𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬‬
‫⟩ ‪𝑎𝑎⃑ + 𝑏𝑏�⃑ = ⟨𝑎𝑎1 + 𝑏𝑏1 , 𝑎𝑎2 + 𝑏𝑏2 , 𝑎𝑎3 + 𝑏𝑏3‬‬
‫ﺃﻣﺎ ﺍﻟﺗﻔﺳﻳﺭ ﺍﻟﻬﻧﺩﺳﻲ ﻟﺣﺎﺻﻝ ﺟﻣﻊ ﻣﺗﺟﻬﻳﻥ ﻧﻭﺿﺣﻪ ﺑﺎﻟﺭﺳﻡ ﺍﻵﺗﻲ ﻭ ﺍﻟﺫﻱ ﻳﻁﻠﻕ ﻋﻠﻳﻪ ﺃﺣﻳﺎﻧﺎ ﻗﺎﻧﻭﻥ‬
‫ﺍﻟﻣﺛﻠﺙ ﺃﻭ ﻗﺎﻧﻭﻥ ﻣﺗﻭﺍﺯﻱ ﺍﻷﺿﻼﻉ ‪.triangle law or parallelogram law‬‬
‫‪3‬‬
‫ﺣﺳﺎﺑﻳﺎ ً ﻳﻛﻭﻥ ﻁﺭﺡ ‪ subtraction‬ﻣﺗﺟﻬﻳﻥ ﻣﺷﺎﺑﻬﺎ ً ﻟﺟﻣﻌﻬﻣﺎ‪.‬‬
‫⟩ ‪𝑎𝑎⃑ − 𝑏𝑏�⃑ = ⟨𝑎𝑎1 − 𝑏𝑏1 , 𝑎𝑎2 − 𝑏𝑏2 , 𝑎𝑎3 − 𝑏𝑏3‬‬
‫ﻭ ﻳﻛﻭﻥ‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﺭﺑﻣﺎ ﻧﺗﺳﺎءﻝ ﻟﻣﺎﺫﺍ ﻳﻛﻭﻥ ﺍﻟﺗﻔﺳﻳﺭ ﺍﻟﻬﻧﺩﺳﻲ ﻟﻁﺭﺡ ﻣﺗﺟﻬﻳﻥ ﺑﻬﺫﻩ ﺍﻟﺻﻭﺭﺓ ؟‬
‫ﺍﻟﺗﻔﺳﻳﺭ ﻛﻣﺎ ﻓﻲ ﺍﻟﺭﺳﻡ ﺍﻟﻣﻭﺿﺢ ﻓﻲ ﺍﻷﺳﻔﻝ ‪ ،‬ﺃﻱ ﻧﺗﻌﺎﻣﻝ ﻣﻊ ⃑�𝑏𝑏 ‪ 𝑎𝑎⃑ −‬ﺑﺎﻟﺻﻭﺭﺓ )⃑�𝑏𝑏 ‪𝑎𝑎⃑ + (−‬‬
‫‪4‬‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﻻ ﻳﻣﻛﻥ ﺃﻥ ﻧﺟﻣﻊ ﺃﻭ ﻧﻁﺭﺡ ﻣﺗﺟﻬﻳﻥ ﻣﺎ ﻟﻡ ﻳﻛﻥ ﻟﻬﻣﺎ ﻧﻔﺱ ﺍﻟﻌﺩﺩ ﻣﻥ ﺍﻟﻣﺭﻛﺑﺎﺕ‪.‬‬
‫ﺃﻣﺎ ﻋﻣﻠﻳﺔ ﺿﺭﺏ ﺍﻟﻣﺗﺟﻪ ﺑﺛﺎﺑﺕ ‪ scalar multiplication‬ﻓﺈﻧﻬﺎ ﺗﻛﻭﻥ ﻛﺂﻻﺗﻲ‪:‬‬
‫ﺃﺫﺍ ﻛﺎﻥ ⟩ ‪ 𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻭ ‪ c‬ﺃﻱ ﺛﺎﺑﺕ ﻓﺎﻥ‪:‬‬
‫⟩ ‪𝑐𝑐𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑐𝑐𝑎𝑎2 , 𝑐𝑐𝑐𝑐3‬‬
‫ﻟﻛﻲ ﻧﺭﻯ ﺍﻟﺗﻣﺛﻳﻝ ﺍﻟﻬﻧﺩﺳﻲ ﻟﺿﺭﺏ ﺍﻟﻣﺗﺟﻪ ﺑﺛﺎﺑﺕ ﺩﻋﻧﺎ ﻧﺄﺧﺫ ﺍﻟﻣﺛﺎﻝ ﺍﻷﺗﻲ‪:‬‬
‫‪1‬‬
‫ﻣﺛﺎﻝ‪ .‬ﻟﻳﻛﻥ ⟩‪ 𝑎𝑎⃑ = ⟨2,4‬ﺍﺣﺳﺏ ⃑𝑎𝑎‪ . 𝑎𝑎⃑ , 3𝑎𝑎⃑, −2‬ﺛﻡ ﺍﺭﺳﻡ ﺗﻠﻙ ﺍﻟﻣﺗﺟﻬﺎﺕ ﻋﻠﻰ ﻧﻔﺱ‬
‫‪2‬‬
‫ﺍﻟﻣﺣﻭﺭﻳﻥ‪.‬‬
‫‪1‬‬
‫⟩‪𝑎𝑎⃑ = ⟨1,2⟩ , − 2𝑎𝑎⃑ = ⟨−4, −8‬‬
‫‪2‬‬
‫‪3𝑎𝑎⃑ = ⟨6,12⟩ ,‬‬
‫ﻣﻥ ﺍﻟﺭﺳﻡ ﺃﻋﻼﻩ ﻣﺎ ﻳﻠﻲ‪:‬‬
‫• ﺃﺫﺍ ﻛﺎﻥ ﺍﻟﺛﺎﺑﺕ ‪ 𝑐𝑐 > 1‬ﻓﺎﻥ ﻁﻭﻝ ﺍﻟﻣﺗﺟﻪ ﺳﻳﺯﻳﺩ‪.‬‬
‫• ﺃﺫﺍ ﻛﺎﻥ ﺍﻟﺛﺎﺑﺕ ‪ 𝑐𝑐 < 1‬ﻓﺎﻥ ﻁﻭﻝ ﺍﻟﻣﺗﺟﻪ ﺳﻳﻘﻝ‪.‬‬
‫• ﺃﺫﺍ ﻛﺎﻥ ﺍﻟﺛﺎﺑﺕ ‪ 𝑐𝑐 < 0‬ﺳﻳﺗﻐﻳﺭ ﺍﺗﺟﺎﻩ ﺍﻟﻣﺗﺟﻪ‪.‬‬
‫ﺗﻌﺭﻳﻑ‪ .‬ﻳﻛﻭﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ ⃑𝑎𝑎 ﻭ ⃑�𝑏𝑏 ﻣﺗﻭﺍﺯﻳﺎﻥ ‪ parallel‬ﺃﺫﺍ ﻭﺟﺩ ‪ c‬ﺑﺣﻳﺙ ﺃﻥ‪:‬‬
‫⃑�𝑏𝑏𝑐𝑐 = ⃑𝑎𝑎‬
‫ﻣﺛﺎﻝ‪ .‬ﺑﻳﻥ ﻓﻳﻣﺎ ﺃﺫﺍ ﻛﺎﻧﺕ ﻣﺟﻣﻭﻋﺔ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺁﻻﺗﻳﺔ ﻣﺗﻭﺍﺯﻳﺔ ﺃﻡ ﻻ‪.‬‬
‫‪5‬‬
‫• ⟩‪ 𝑏𝑏�⃑ = ⟨−6,12, −3‬ﻭ ⟩‪𝑎𝑎⃑ = ⟨2, −4,1‬‬
‫• ⟩‪ 𝑏𝑏�⃑ = ⟨2, −9‬ﻭ ⟩‪𝑎𝑎⃑ = ⟨4,10‬‬
‫ﺍﻟﺣﻝ‪ .a.‬ﺑﻣﺎ ﺃﻥ ⃑𝑎𝑎‪ 𝑏𝑏�⃑ = −3‬ﻓﺎﻥ ﻣﺟﻣﻭﻋﺔ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺍﻻﻭﻟﻰ ﻣﺗﻭﺍﺯﻳﺔ‪.‬‬
‫‪ .b‬ﺃﻥ ﻣﺟﻣﻭﻋﺔ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺍﻟﺛﺎﻧﻳﺔ ﻻ ﺗﻛﻭﻥ ﻣﺗﻭﺍﺯﻳﺔ ﻷﻧﻪ ﻻ ﻳﻣﻛﻥ ﻛﺗﺎﺑﺔ ﺃﺣﺩﺍﻫﺎ ﻣﺳﺎﻭﻳﺎ ً ﺇﻟﻰ‬
‫ﺛﺎﺑﺕ ﻣﺿﺭﻭﺑﺎ ً ﻓﻲ ﺍﻟﻣﺗﺟﻪ ﺍﻷﺧﺭ‪.‬‬
‫𝑤𝑤‪.‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺟﺩ ﻣﺗﺟﻪ ﺍﻟﻭﺣﺩﺓ ﺍﻟﺫﻱ ﻳﻛﻭﻥ ﺑﻧﻔﺱ ﺍﻻﺗﺟﺎﻩ ﻣﻊ ﺍﻟﻣﺗﺟﻪ ⟩‪��⃑ = ⟨−5,2,1‬‬
‫ﺍﻟﺣﻝ‪ .‬ﻻﺣﻅ ﺃﻥ‬
‫ﻣﻥ ﺍﻟﻭﺍﺿﺢ ﺃﻥ‬
‫‪5‬‬
‫𝑤𝑤‬
‫⃑��‬
‫‪1‬‬
‫‪2‬‬
‫‪1‬‬
‫‪⟨−5,2,1⟩ = ⟨−‬‬
‫⟩‬
‫=‬
‫‪,‬‬
‫‪,‬‬
‫𝑤𝑤‖‬
‫‪��⃑‖ √30‬‬
‫‪√30 √30 √30‬‬
‫‪1‬‬
‫‪2‬‬
‫‪√30‬‬
‫‪= 1.‬‬
‫= �‬
‫‪√30‬‬
‫‪√30 √30 √30‬‬
‫𝑢𝑢 ﻭ ﻫﻭ ﻣﺗﺟﻪ ﻭﺣﺩﺓ‪.‬‬
‫𝑤𝑤 ﻷﻧﻪ ﻳﺳﺎﻭﻱ ﺛﺎﺑﺕ ﻣﺿﺭﻭﺏ ﺑـ ⃑�‬
‫𝑢𝑢 ﺑﻧﻔﺱ ﺍﻻﺗﺟﺎﻩ ﻣﻊ ⃑��‬
‫ﺃﻥ ﺍﻟﻣﺗﺟﻪ ⃑�‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﺑﺻﻭﺭﺓ ﻋﺎﻣﺔ ﺃﺫﺍ ﻛﺎﻥ ﻟﺩﻳﻧﺎ‬
‫ﺍﻻﺗﺟﺎﻩ ﻣﻊ ⃑��‬
‫𝑤𝑤‪.‬‬
‫‪1‬‬
‫‪+‬‬
‫‪4‬‬
‫𝑢𝑢‬
‫= ⃑�‬
‫‪+‬‬
‫‪25‬‬
‫𝑢𝑢‖‬
‫� = ‖⃑�‬
‫⃑��‬
‫𝑤𝑤‬
‫𝑢𝑢 ﻫﻭ ﻣﺗﺟﻪ ﺍﻟﻭﺣﺩﺓ ﺍﻟﺫﻱ ﻳﻛﻭﻥ ﺑﻧﻔﺱ‬
‫𝑤𝑤 ﻓﺎﻥ ‖⃑��𝑤𝑤‖ = ⃑�‬
‫⃑��‬
‫ﻣﺗﺟﻬﺎﺕ ﺍﻟﻘﺎﻋﺩﺓ ﺍﻷﺳﺎﺱ ‪standard basis vectors‬‬
‫ﻟﻳﻛﻥ ⟩ ‪ 𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻣﺗﺟﻬﺎ ً ﻓﺎﻧﻧﺎ ﻳﻣﻛﻥ ﺃﻥ ﻧﻛﺗﺑﻪ ﺑﺎﻟﺻﻭﺭﺓ‪:‬‬
‫⟩ ‪𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3 ⟩ = ⟨𝑎𝑎1 , 0,0⟩ + ⟨0, 𝑎𝑎2 , 0⟩ + ⟨0,0, 𝑎𝑎3‬‬
‫⟩‪= 𝑎𝑎1 ⟨1,0,0⟩ + 𝑎𝑎2 ⟨0,1,0⟩ +, 𝑎𝑎3 ⟨0,0,1‬‬
‫ﻻﺣﻅ ﺃﻥ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺍﻟﺛﻼﺛﺔ ﺍﻷﺧﻳﺭﺓ ﻫﻲ ﻣﺗﺟﻬﺎﺕ ﺍﻟﻘﺎﻋﺩﺓ ﺍﻷﺳﺎﺱ ﻟﻠﻔﺿﺎء ﺍﻟﺛﻼﺛﻲ ﺍﻷﺑﻌﺎﺩ‪ .‬ﻭ ﺗﻛﺗﺏ‬
‫⃑�‬
‫𝑘𝑘 ‪𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3 ⟩ = 𝑎𝑎1 𝑖𝑖⃑ + 𝑎𝑎2 𝑗𝑗⃑ +, 𝑎𝑎3‬‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﻳﻣﻛﻥ ﻛﺗﺎﺑﺔ ﺃﻱ ﻣﺗﺟﻪ ﺑﺎﻟﺻﻭﺭﺓ ﺃﻋﻼﻩ‪.‬‬
‫ﺗﻣﺭﻳﻥ‪ .‬ﺃﺫﺍ ﻛﺎﻥ ⟩‪ 𝑎𝑎⃑ = ⟨3, −9,1‬ﻭ ⃑�‬
‫𝑤𝑤‪.2𝑎𝑎⃑ − 3‬‬
‫𝑤𝑤 ﺃﺣﺳﺏ ⃑��‬
‫𝑘𝑘‪��⃑ = −𝑖𝑖⃑ + 8‬‬
‫‪6‬‬
‫ﺃﻻﻥ ﺩﻋﻧﺎ ﻧﻘﺩﻡ ﺑﻌﺽ ﺧﺻﺎﺋﺹ ﺟﺑﺭ ﺍﻟﻣﺗﺟﻬﺎﺕ‪.‬‬
‫ﺧﺻﺎﺋﺹ‪.‬‬
‫‪ 𝑣𝑣,‬ﻣﺗﺟﻬﺎﺕ ﻭ ﻟﻳﻛﻥ ‪ a ,b‬ﺃﻋﺩﺍﺩ ﻓﺎﻥ‪:‬‬
‫‪���⃑ 𝑤𝑤,‬‬
‫𝑢𝑢 ⃑����‬
‫ﻟﻳﻛﻥ ﻛﻝ ﻣﻥ ⃑�‬
‫𝑤𝑤 ‪𝑣𝑣⃑ +‬‬
‫𝑤𝑤 = ⃑��‬
‫⃑𝑣𝑣 ‪��⃑ +‬‬
‫⃑𝑣𝑣 = ‪𝑣𝑣⃑ + 0‬‬
‫⃑��‬
‫𝑤𝑤 ‪𝑎𝑎(𝑣𝑣⃑ +‬‬
‫𝑤𝑤𝑎𝑎 ‪��⃑) = 𝑎𝑎𝑣𝑣⃑ +‬‬
‫𝑤𝑤 ‪�⃑ + 𝑣𝑣⃑) +‬‬
‫𝑢𝑢‬
‫𝑤𝑤 ‪�⃑ + (𝑣𝑣⃑ +‬‬
‫𝑢𝑢( = )⃑��‬
‫⃑��‬
‫⃑𝑣𝑣 = ⃑𝑣𝑣‪1‬‬
‫⃑𝑣𝑣𝑏𝑏 ‪(𝑎𝑎 + 𝑏𝑏)𝑣𝑣⃑ = 𝑎𝑎𝑣𝑣⃑ +‬‬
‫•‬
‫•‬
‫•‬
‫•‬
‫•‬
‫•‬
‫ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ ﺃﻭ ﺍﻟﻧﻘﻁﻲ ﺃﻭ ﺍﻟﺩﺍﺧﻠﻲ ‪scalar or dot or inner product‬‬
‫ﺗﻌﺭﻳﻑ‪ .‬ﻟﻳﻛﻥ ﻛﻝ ﻣﻥ ⟩ ‪ 𝑏𝑏�⃑ = ⟨𝑏𝑏1 , 𝑏𝑏2 , 𝑏𝑏3‬ﻭ ⟩ ‪ 𝑎𝑎⃑ = ⟨𝑎𝑎1 , 𝑎𝑎2 , 𝑎𝑎3‬ﻣﺗﺟﻪ‪ .‬ﻓﺎﻥ ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ‬
‫ﺃﻭ ﺍﻟﻧﻘﻁﻲ ﺃﻭ ﺍﻟﺩﺍﺧﻠﻲ ﻟﻬﻣﺎ ﻫﻭ‪:‬‬
‫‪𝑎𝑎⃑. 𝑏𝑏�⃑ = 𝑎𝑎1 𝑏𝑏1 + 𝑎𝑎2 𝑏𝑏2 + 𝑎𝑎3 𝑏𝑏3‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺍﺣﺳﺏ ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ ﻟﻛﻝ ﻣﻣﺎ ﻳﺄﺗﻲ‪:‬‬
‫𝑤𝑤 ﻭ ⃑𝑗𝑗‪𝑣𝑣⃑ = 5𝑖𝑖⃑ − 8‬‬
‫• ⃑𝑗𝑗‪��⃑ = 𝑖𝑖⃑ + 2‬‬
‫• ⟩‪ 𝑏𝑏�⃑ = ⟨2,3,1‬ﻭ ⟩‪𝑎𝑎⃑ = ⟨0,3, −7‬‬
‫ﺍﻟﺣﻝ‪.‬‬
‫ﺃﻻﻥ ﺩﻋﻧﺎ ﻧﻘﺩﻡ ﺑﻌﺽ ﺧﺻﺎﺋﺹ ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ‪.‬‬
‫ﺧﺻﺎﺋﺹ‪.‬‬
‫‪ 𝑣𝑣,‬ﻣﺗﺟﻬﺎﺕ ﻓﺎﻥ‪:‬‬
‫‪���⃑ 𝑤𝑤,‬‬
‫𝑢𝑢 ⃑����‬
‫ﻟﻳﻛﻥ ﻛﻝ ﻣﻥ ⃑�‬
‫• ﺃﺫﺍ ﻛﺎﻥ ‪ 𝑣𝑣⃑. 𝑣𝑣⃑ = 0‬ﻓﺎﻥ ‪.𝑣𝑣⃑ = 0‬‬
‫‪7‬‬
‫𝑤𝑤 ‪• 𝑣𝑣⃑.‬‬
‫‪��⃑ = 5 − 16 = −11‬‬
‫‪• 𝑎𝑎⃑. 𝑏𝑏�⃑ = 0 + 9 − 7 = 2‬‬
‫𝑢𝑢 ‪�⃑. 𝑣𝑣⃑ +‬‬
‫𝑢𝑢‬
‫𝑤𝑤 ‪�⃑. (𝑣𝑣⃑ +‬‬
‫𝑢𝑢 = )⃑��‬
‫𝑤𝑤 ‪�⃑.‬‬
‫⃑��‬
‫𝑤𝑤 ‪𝑣𝑣⃑.‬‬
‫𝑤𝑤 = ⃑��‬
‫⃑𝑣𝑣 ‪��⃑.‬‬
‫‪𝑣𝑣⃑. 𝑣𝑣⃑ = ‖𝑣𝑣⃑‖2‬‬
‫𝑤𝑤 ‪(𝑐𝑐𝑣𝑣⃑).‬‬
‫)⃑��‬
‫𝑤𝑤𝑐𝑐( ‪��⃑ = 𝑣𝑣⃑.‬‬
‫𝑤𝑤 ‪��⃑) = 𝑐𝑐(𝑣𝑣⃑.‬‬
‫•‬
‫•‬
‫•‬
‫•‬
‫⃑����ﻭ ⃑𝑎𝑎 ﺑﺣﻳﺙ‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﻳﻭﺟﺩ ﺗﻣﺛﻳﻝ ﻫﻧﺩﺳﻲ ﻟﻠﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ‪ .‬ﻟﺗﻛﻥ ‪ θ‬ﻫﻲ ﺍﻟﺯﺍﻭﻳﺔ ﺑﻳﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ 𝑏𝑏‬
‫ﺃﻥ ‪ 0 ≤ θ ≤ π‬ﻛﻣﺎ ﻓﻲ ﺍﻟﺭﺳﻡ ﺃﺩﻧﺎﻩ‪:‬‬
‫ﻣﻼﺣﻅﺔ‪.‬‬
‫• 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐�⃑�𝑏𝑏�‖⃑𝑎𝑎‖ = ⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫• ﺃﻥ ﺻﻳﻐﺔ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻘﻁﻲ ﻓﻲ ﺍﻟﺻﻳﻐﺔ ﺃﻋﻼﻩ ﻻ ﺗﺳﺗﺧﺩﻡ ﻻﺳﺗﺧﺭﺍﺝ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻘﻁﻲ‬
‫ﻟﻣﺗﺟﻬﻳﻥ ﻭ ﺃﻧﻣﺎ ﺗﺳﺗﺧﺩﻡ ﻻﺳﺗﺧﺭﺍﺝ ﺍﻟﺯﺍﻭﻳﺔ ﺍﻟﻣﺣﺻﻭﺭﺓ ﺑﻳﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ‪.‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺟﺩ ﺍﻟﺯﺍﻭﻳﺔ ﺑﻳﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ ⟩‪ 𝑎𝑎⃑ = ⟨3, −4, −1‬ﻭ ⟩‪. 𝑏𝑏�⃑ = ⟨0,5,2‬‬
‫ﺍﻟﺣﻝ‪ .‬ﻧﺣﺗﺎﺝ ﺇﻟﻰ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻔﻁﻲ ﻻﺳﺗﺧﺭﺍﺝ ﺍﻟﺯﺍﻭﻳﺔ ﺑﻳﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ‬
‫ﻟﺩﻳﻧﺎ‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﺗﻛﻭﻥ‬
‫‪𝑎𝑎⃑. 𝑏𝑏�⃑ = −22, ‖𝑎𝑎⃑‖ = √26, �𝑏𝑏�⃑� = √29‬‬
‫⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫‪−22‬‬
‫=‬
‫‪= −0.8011927‬‬
‫‪‖𝑎𝑎⃑‖�𝑏𝑏�⃑� √26√29‬‬
‫= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐‬
‫‪𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑐𝑐 −1 (−0.8011927) = 2.5 radians.‬‬
‫ﻣﻼﺣﻅﺔ‪ .‬ﻳﻌﻁﻳﻧﺎ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻘﻁﻲ ﻁﺭﻳﻘﺔ ﻟﻣﻌﺭﻓﺔ ﻓﻳﻣﺎ ﺃﺫﺍ ﻛﺎﻥ ﻣﺗﺟﻬﻳﻥ ﻣﺗﻌﺎﻣﺩﻳﻥ‬
‫‪ orthogonal‬ﺑﺩﻻً ﻣﻥ‬
‫‪ perpendicular‬ﺃﻡ ﻻ‪ .‬ﻭ ﺳﻧﺳﺗﺧﺩﻡ ﺍﻟﻣﺻﻁﻠﺢ‬
‫‪8‬‬
‫‪ .perpendicular‬ﻛﻣﺎ ﺳﻳﻌﻁﻳﻧﺎ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻘﻁﻲ ﺃﻳﺿﺎ ﻁﺭﻳﻘﺔ ﻟﻣﻌﺭﻓﺔ ﻓﻳﻣﺎ ﺃﺫﺍ ﻛﺎﻥ ﻣﺗﺟﻬﻳﻥ‬
‫ﻣﺗﻭﺍﺯﻳﻳﻥ ‪ parallel‬ﺃﻡ ﻻ‪.‬‬
‫ﺃﻻﻥ ﺃﺫﺍ ﻛﺎﻥ ﺍﻟﻣﺗﺟﻬﺎﻥ ﻣﺗﻌﺎﻣﺩﻳﻥ ﻧﺣﻥ ﻧﻌﻠﻡ ﺃﻥ ﺍﻟﺯﺍﻭﻳﺔ ﺑﻳﻧﻬﻣﺎ ﺗﺳﺎﻭﻱ ∘‪ 90‬ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﻧﺣﺻﻝ‬
‫ﻣﻥ ﺍﻟﻣﻼﺣﻅﺔ ﺍﻋﻼﻩ ﺍﻥ ‪ 𝑎𝑎⃑. 𝑏𝑏�⃑ = 0‬ﻭ ﻧﺣﻥ ﻧﻌﻠﻡ ﺍﻳﺿﺎ ً ﺍﺫﺍ ﻛﺎﻥ ﻣﺗﺟﻬﺎﻥ ﻣﺗﻭﺍﺯﻳﻳﻥ ﻓﺎﻥ ﺍﻟﺯﺍﻭﻳﺔ‬
‫∘‪ .180‬ﺃﻱ‬
‫ﺑﻳﻧﻬﻣﺎ ﺃﻣﺎ ﺗﺳﺎﻭﻱ ﺻﻔﺭ ﺃﻭ ﻳﻛﻭﻧﺎﻥ ﺑﻧﻔﺱ ﺍﻻﺗﺟﺎﻩ ﺃﻭ ﺗﻛﻭﻥ ﺍﻟﺯﺍﻭﻳﺔ ﺑﻳﻧﻬﻣﺎ ﺗﺳﺎﻭﻱ‬
‫ﻳﻛﻭﻧﺎﻥ ﺑﺎﺗﺟﺎﻫﻳﻥ ﻣﺗﻌﺎﻛﺳﻳﻥ ﻭ ﺑﺎﺳﺗﺧﺩﺍﻡ ﺍﻟﻣﻼﺣﻅﺔ ﺍﻋﻼﻩ ﻣﺭﺓ ﺍﺧﺭﻯ ﺳﻧﺣﺻﻝ ﻋﻠﻰ‪:‬‬
‫‪ 𝜃𝜃 = 0‬ﻋﻧﺩﻣﺎ �⃑�𝑏𝑏�‖⃑𝑎𝑎‖ = ⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫∘‪ 𝜃𝜃 = 180‬ﻋﻧﺩﻣﺎ �⃑�𝑏𝑏�‖⃑𝑎𝑎‖‪𝑎𝑎⃑. 𝑏𝑏�⃑ = −‬‬
‫ﻣﺛﺎﻝ‪ .‬ﺑﻳﻥ ﻓﻳﻣﺎ ﺃﺫﺍ ﻛﺎﻧﺕ ﺍﻟﻣﺗﺟﻬﺎﺕ ﺍﻵﺗﻳﺔ ﻣﺗﻭﺍﺯﻳﺔ ﺃﻡ ﻣﺗﻌﺎﻣﺩﺓ ﺃﻡ ﻻ ﻣﺗﻌﺎﻣﺩﺓ ﻭ ﻻ ﻣﺗﻭﺍﺯﻳﺔ‪.‬‬
‫• ⟩‪ 𝑏𝑏�⃑ = ⟨2,5,2‬ﻭ ⟩‪𝑎𝑎⃑ = ⟨6, −2, −1‬‬
‫‪1 1‬‬
‫𝑢𝑢‬
‫• ⟩ ‪ 𝑣𝑣⃑ = ⟨ ,‬ﻭ ⟩‪�⃑ = ⟨2, −1‬‬
‫‪2 4‬‬
‫ﺍﻟﺣﻝ‪.‬‬
‫• ﻓﻲ ﺍﻟﺑﺩﺍﻳﺔ ﺩﻋﻧﺎ ﻧﺭﻯ ﻓﻳﻣﺎ ﺃﺫﺍ ﻛﺎﻥ ﺍﻟﻣﺗﺟﻬﺎﻥ ﻣﺗﻌﺎﻣﺩﻳﻥ ﺃﻡ ﻻ‬
‫‪𝑎𝑎⃑. 𝑏𝑏�⃑ = 12 − 10 − 2 = 0‬‬
‫⃑��ﻭ⃑𝑎𝑎 ﻣﺗﻌﺎﻣﺩﺍﻥ‪.‬‬
‫ﻭ ﻫﺫﺍ ﻳﺅﺩﻱ ﺇﻟﻰ ﺃﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ 𝑏𝑏‬
‫ﻛﺫﻟﻙ ﺩﻋﻧﺎ ﻧﻼﺣﻅ ﺍﻟﺿﺭﺏ ﺍﻟﻧﻘﻁﻲ ﻟﻠﻣﺗﺟﻬﻳﻥ‬
‫‪1‬‬
‫‪5‬‬
‫‪=−‬‬
‫‪4‬‬
‫‪4‬‬
‫𝑢𝑢‬
‫‪�⃑. 𝑣𝑣⃑ = −1 −‬‬
‫ﺳﻧﺣﺻﻝ ﻣﻥ ﻫﺫﺍ ﺃﻥ ﺍﻟﻣﺗﺟﻬﻳﻥ ﻏﻳﺭ ﻣﺗﻌﺎﻣﺩﻳﻥ‪ .‬ﻭ ﺃﻻﻥ ﺩﻋﻧﺎ ﻧﺳﺗﺧﺭﺝ ﻁﻭﻝ ﻛﻝ ﻣﻧﻬﻣﺎ‬
‫ﻭ ﺃﻻﻥ ﻻﺣﻅ ﺃﻥ‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﻳﻛﻭﻥ ﺍﻟﻣﺗﺟﻬﺎﻥ ﻣﺗﻭﺍﺯﻳﻳﻥ‪.‬‬
‫‪5‬‬
‫‪√5‬‬
‫𝑢𝑢‖‬
‫= � = ‖⃑𝑣𝑣‖ ‪�⃑‖ = √5,‬‬
‫‪4‬‬
‫‪16‬‬
‫‪5‬‬
‫‪√5‬‬
‫‖⃑𝑣𝑣‖‖⃑�‬
‫𝑢𝑢‬
‫𝑢𝑢‖‪�⃑. 𝑣𝑣⃑ = − = −√5 � � = −‬‬
‫‪4‬‬
‫‪4‬‬
‫ﺃﻥ ﺗﻁﺑﻳﻘﺎﺕ ﺍﻟﺿﺭﺏ ﺍﻟﻌﺩﺩﻱ ﻫﻲ ﺍﻟﻣﺳﺎﻗﻁ ‪.projections‬‬
‫‪9‬‬
‫ﺍﻟﻣﺳﺎﻗﻁ ‪projections‬‬
‫⃑��ﻭ⃑𝑎𝑎 ﻣﺗﺟﻬﻳﻥ‪ .‬ﺳﻧﺷﺭﺡ ﻣﻌﻧﻰ ﻣﺳﻘﻁ ⃑�𝑏𝑏 ﻋﻠﻰ ⃑𝑎𝑎 ﺍﻟﺫﻱ ﺳﻧﺭﻣﺯ ﻟﻪ ﺑﺎﻟﺭﻣﺯ‬
‫ﻟﻳﻛﻥ ﻛﻝ ﻣﻥ 𝑏𝑏‬
‫⃑�𝑏𝑏 ⃑�𝑎𝑎𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ‪.‬‬
‫ﺗﻭﺟﺩ ﺻﻳﻐﺔ ﻹﻳﺟﺎﺩ ﻣﺳﻘﻁ ⃑�𝑏𝑏 ﻋﻠﻰ ⃑𝑎𝑎‬
‫⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫⃑𝑎𝑎‬
‫‪‖𝑎𝑎⃑‖2‬‬
‫ﺍﻣﺎ ﻣﺳﻘﻁ ⃑𝑎𝑎 ﻋﻠﻰ ⃑�𝑏𝑏‬
‫⃑�𝑏𝑏‬
‫⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫‪2‬‬
‫�⃑�𝑏𝑏�‬
‫= ⃑�𝑏𝑏 ⃑�𝑎𝑎𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝‬
‫= ⃑𝑎𝑎 ⃑�𝑏𝑏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝‬
‫ﻣﺛﺎﻝ‪ .‬ﺟﺩ ﻣﺳﻘﻁ ⟩‪ 𝑏𝑏�⃑ = ⟨2,1, −1‬ﻋﻠﻰ ﻭ ⟩‪.𝑎𝑎⃑ = ⟨1,0, −2‬‬
‫ﺍﻟﺣﻝ‪ .‬ﻻﺣﻅ ﺃﻥ‬
‫‪𝑎𝑎⃑. 𝑏𝑏�⃑ = 4 , ‖𝑎𝑎⃑‖2 = 5‬‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﻳﻛﻭﻥ‬
‫‪10‬‬
‫‪8‬‬
‫‪4‬‬
‫⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫⃑𝑎𝑎‬
‫‪‖𝑎𝑎⃑‖2‬‬
‫‪4‬‬
‫= ⃑�𝑏𝑏 ⃑�𝑎𝑎𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝‬
‫‪= ⟨1,0, −2⟩ = ⟨ , 0, − ⟩.‬‬
‫‪5‬‬
‫‪5‬‬
‫‪5‬‬
‫ﺃﻻﻥ ﺩﻋﻧﺎ ﻧﺳﺗﺑﺩﻝ ﺍﻟﻣﺗﺟﻪ ⃑𝑎𝑎 ﺑﺎﻟﻣﺗﺟﻪ ⃑�𝑏𝑏‪ .‬ﻣﺎﺫﺍ ﺳﻧﻼﺣﻅ ؟!‬
‫‪2‬‬
‫‪𝑎𝑎⃑. 𝑏𝑏�⃑ = 4 , �𝑏𝑏�⃑� = 6‬‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﻳﻛﻭﻥ‬
‫‪2‬‬
‫⃑�𝑏𝑏‬
‫‪4 2‬‬
‫⃑�𝑏𝑏 ‪𝑎𝑎⃑.‬‬
‫‪2‬‬
‫�⃑�𝑏𝑏�‬
‫‪4‬‬
‫= ⃑𝑎𝑎 ⃑�𝑏𝑏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝‬
‫‪= ⟨2,1, −1⟩ = ⟨ , , − ⟩.‬‬
‫‪3‬‬
‫‪6‬‬
‫‪3 3‬‬
‫ﻻﺣﻅ ﺃﻥ ﻛﻼ ﺍﻟﻣﺳﻘﻁﻳﻥ ﻣﺧﺗﻠﻔﻳﻥ‪.‬‬
‫ﺗﻣﺎﺭﻳﻥ ﻣﺣﻠﻭﻟﺔ‪.‬‬
‫ﻣﻼﺣﻅﺔ ‪ .‬ﺩﻋﻧﺎ ﻧﺗﺫﻛﺭ ﻓﻲ ﺍﻟﺑﺩﺍﻳﺔ ﺍﻟﺗﻣﺛﻳﻝ ﺍﻟﻬﻧﺩﺳﻲ ﻟﺣﺎﺻﻝ ﺟﻣﻊ ﺍﻟﻣﺗﺟﻬﻳﻥ ⃑�‬
‫𝐵𝐵 ﻭ ⃑𝐴𝐴‪.‬‬
‫⃑�‬
‫𝐵𝐵‬
‫⃑�‬
‫𝐵𝐵‪𝐴𝐴⃑+‬‬
‫⃑𝐴𝐴‬
‫⃑�‬
‫𝐵𝐵‬
‫⃑𝐴𝐴‬
‫‪.1‬ﺑﺭﻫﻥ ﺃﻥ ﺍﻟﺧﻁ ﺍﻟﺫﻱ ﻳﺻﻝ ﺑﻳﻥ ﻧﺻﻔﻲ ﺿﻠﻌﻳﻥ ﻓﻲ ﻣﺛﻠﺙ ﻳﻭﺍﺯﻱ ﺿﻠﻌﻪ ﺍﻟﺛﺎﻟﺙ ﻭ ﻳﺳﺎﻭﻱ ﻧﺻﻔﻪ‬
‫ﺑﺎﻟﻘﻳﺎﺱ‪.‬‬
‫‪�⃑ = C‬‬
‫⃑�‬
‫‪�⃑ + A‬‬
‫‪ . B‬ﺍﻥ ﺍﻟﻣﺗﺟﻪ ⃑��‬
‫‪ D‬ﻳﺻﻝ ﺑﻳﻥ ﻣﻧﺗﺻﻔﻲ‬
‫ﺍﻟﺑﺭﻫﺎﻥ‪ .‬ﻣﻥ ﺍﻟﺷﻛﻝ ﻓﻲ ﺍﻷﺳﻔﻝ ﻟﺩﻳﻧﺎ‬
‫ﺍﻟﺿﻠﻌﻳﻥ ⃑�‬
‫𝐵𝐵 ﻭ ⃑𝐴𝐴 ‪ .‬ﻓﺎﻥ‬
‫‪�⃑ = 1 B‬‬
‫‪�D‬‬
‫⃑���� ‪�⃑ + 1 �A⃑ = 1‬‬
‫‪�⃑) = 1 �C⃑.‬‬
‫𝐵𝐵‪(𝐴𝐴+‬‬
‫‪2‬‬
‫‪2‬‬
‫‪2‬‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﻳﻛﻭﻥ ⃑�‬
‫‪ �D‬ﻣﻭﺍﺯﻱ ﺍﻟﻰ ⃑‪ �C‬ﻭ ﻳﺳﺎﻭﻱ ﻧﺻﻔﻪ ﺑﺎﻟﻘﻳﺎﺱ‪.‬‬
‫𝐵𝐵‬
‫‪11‬‬
‫𝐸𝐸‬
‫⃑𝐴𝐴‬
‫⃑𝐶𝐶‬
‫𝐶𝐶‬
‫⃑�‬
‫𝐷𝐷‬
‫𝐷𝐷‬
‫⃑�‬
‫𝐵𝐵‬
‫‪2‬‬
‫𝐴𝐴‬
‫‪ .2‬ﺟﺩ ﺍﻟﺯﺍﻭﻳﺔ 𝐵𝐵𝐵𝐵𝐵𝐵 ≺=∝ ﻟﻠﻣﺛﻠﺙ 𝐴𝐴𝐴𝐴𝐴𝐴 ﺍﻟﺫﻱ ﺭﺅﻭﺳﻪ‬
‫)‪𝐴𝐴 = (1,0,1) , 𝐵𝐵 = (2, −1,1) , 𝐶𝐶 = (−2,1,0‬‬
‫𝐶𝐶‬
‫⃑𝐴𝐴‬
‫ﺍﻟﺣﻝ‪.‬‬
‫⃑�‬
‫𝐵𝐵‬
‫𝐴𝐴‬
‫𝐵𝐵‬
‫⟩‪𝐴𝐴⃑ = 𝐴𝐴𝐴𝐴 = ⟨−3,1, −1‬‬
‫ﻟﺩﻳﻧﺎ‬
‫⟩‪�⃑ = 𝐴𝐴𝐴𝐴 = ⟨1, −1,0‬‬
‫𝐵𝐵‬
‫⃑�‬
‫)‪(−3 − 1‬‬
‫𝐵𝐵 ‪𝐴𝐴⃑.‬‬
‫=‬
‫‪= −0.85280.‬‬
‫�⃑�‬
‫‪√22‬‬
‫𝐵𝐵��⃑𝐴𝐴�‬
‫ﻭ ﺑﺎﻟﺗﺎﻟﻲ ﺳﺗﻛﻭﻥ ́‪.∝= 148∘ . 51‬‬
‫‪12‬‬
‫=∝ 𝑐𝑐𝑐𝑐𝑐𝑐‬