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Last updated at Feb. 1, 2020 by Teachoo
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Misc 12 Let ๐ โ = ๐ ฬ + 4๐ ฬ + 2๐ ฬ, ๐ โ = 3๐ ฬ โ 2๐ ฬ + 7๐ ฬ and ๐ โ = 2๐ ฬ โ ๐ ฬ + 4๐ ฬ . Find a vector ๐ โ which is perpendicular to both ๐ โ and ๐ โ and ๐ โ โ ๐ โ = 15 . ๐ โ = ๐ ฬ + 4๐ ฬ + 2๐ ฬ = 1๐ ฬ + 4๐ ฬ + 2๐ ฬ ๐ โ = 3๐ ฬ + 2๐ ฬ + 7๐ ฬ ๐ โ = 2๐ ฬ + ๐ ฬ + 4๐ ฬ = 2๐ ฬ โ 1๐ ฬ + 4๐ ฬ Let ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ Since ๐ โ is perpendicular to ๐ โ and ๐ โ ๐ โ . ๐ โ = 0 & ๐ โ . ๐ โ = 0 ๐ โ . ๐ โ = 0 (x๐ ฬ + y๐ ฬ + z๐ ฬ). (1๐ ฬ + 4๐ ฬ + 2๐ ฬ) = 0 (x ร 1) + (y ร 4) + (z ร 2) = 0 x + 4y + 2z = 0 ๐ โ . ๐ โ = 0 (x๐ ฬ + y๐ ฬ + z๐ ฬ). (3๐ ฬ โ 2๐ ฬ + 7๐ ฬ) = 0 (x ร 3) + (y ร -2) + (z ร 7) = 0 3x โ 2y + 7z = 0 Also, ๐ โ . ๐ โ = 15 (2๐ ฬ โ 1๐ ฬ + 4๐ ฬ). (x๐ ฬ + y๐ ฬ + z๐ ฬ) = 15 (2 ร x) + (โ1 ร y) + (4 ร 2) = 15 2x โ y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 3x โ 2y + 7z = 0 & 2x โ y + 4z = 15 Solving x + 4y + 2z = 0 3x โ 2y + 7z = 0 ๐ฅ/(28 โ (โ4) ) = ๐ฆ/(6 โ 7 ) = ๐ง/(โ2 โ 12 ) ๐ฅ/(32 ) = ๐ฆ/(โ1 ) = ๐ง/(โ14 ) Writing x & y in terms of z โด x = 32๐ง/(โ14) = (โ16๐ง)/7 & y = (โ1๐ง)/(โ14) = ๐ง/14 Putting values of x and y in (3) 2x โ y + 4z = 15 2 ((โ16๐ง)/7 ) โ (๐ง/14 ) + 4z = 5 (โ32)/7 z โ 1/14 z + 4/1 z = 15 ((โ64 โ 1 + 56))/14 z = 15 (โ9)/14 z = 15 z = 15 ร 14/(โ9) z = (โ๐๐)/๐ Putting value of z in x & y, x = (โ16๐ง)/7 = (โ16)/7 ร (โ70)/3 = ๐๐๐/๐ y = ๐ง/14 = 1/14 ร (โ70)/3 = (โ๐)/๐ Therefore, the required vector ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ = 160/3 ๐ ฬ โ 5/3 ๐ ฬ โ 70/3 ๐ ฬ = ๐/๐ (160๐ ฬ โ 5๐ ฬ โ 70๐ ฬ) Note: Answer given in the book is incorrect If we have made any mistake, please comment below
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