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Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 3
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 4
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 5

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Transcript

Misc 12 Let 𝑎 ⃗ = 𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂, 𝑏 ⃗ = 3𝑖 ̂ − 2𝑗 ̂ + 7𝑘 ̂ and 𝑐 ⃗ = 2𝑖 ̂ − 𝑗 ̂ + 4𝑘 ̂ . Find a vector 𝑑 ⃗ which is perpendicular to both 𝑎 ⃗ and 𝑏 ⃗ and 𝑐 ⃗ ⋅ 𝑑 ⃗ = 15 . Given 𝒂 ⃗ = 𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂ 𝒃 ⃗ = 3𝑖 ̂ - 2𝑗 ̂ + 7𝑘 ̂ 𝒄 ⃗ = 2𝑖 ̂ + 𝑗 ̂ + 4𝑘 ̂ Let 𝒅 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ Since 𝒅 ⃗ is perpendicular to 𝑎 ⃗ and 𝑏 ⃗ 𝑑 ⃗ . 𝑎 ⃗ = 0 & 𝑑 ⃗ . 𝑏 ⃗ = 0 𝒅 ⃗ . 𝒂 ⃗ = 0 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). (1𝑖 ̂ + 4𝑗 ̂ + 2𝑘 ̂) = 0 (x × 1) + (y × 4) + (z × 2) = 0 x + 4y + 2z = 0 𝒅 ⃗ . 𝒃 ⃗ = 0 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). (3𝑖 ̂ − 2𝑗 ̂ + 7𝑘 ̂) = 0 (x × 3) + (y × -2) + (z × 7) = 0 3x − 2y + 7z = 0 Also, 𝑐 ⃗ . 𝑑 ⃗ = 15 (2𝑖 ̂ – 1𝑗 ̂ + 4𝑘 ̂). (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂) = 15 (2 × x) + (−1 × y) + (4 × 2) = 15 2x − y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 3x − 2y + 7z = 0 & 2x − y + 4z = 15 Solving x + 4y + 2z = 0 3x – 2y + 7z = 0 𝑥/(28 − (−4) ) = 𝑦/(6 − 7 ) = 𝑧/(−2 − 12 ) 𝒙/(𝟑𝟐 ) = 𝒚/(−𝟏 ) = 𝒛/(−𝟏𝟒 ) Writing x & y in terms of z ∴ x = 32𝑧/(−14) = (−𝟏𝟔𝒛)/𝟕 & y = (−1𝑧)/(−14) = 𝒛/𝟏𝟒 Putting values of x and y in (3) 2x – y + 4z = 15 2 ((−16𝑧)/7 ) − (𝑧/14 ) + 4z = 5 (−32)/7 z − 1/14 z + 4/1 z = 15 ((−64 − 1 + 56))/14 z = 15 (−9)/14 z = 15 z = 15 × 14/(−9) z = (−𝟕𝟎)/𝟑 Putting value of z in x & y, x = (−16𝑧)/7 = (−16)/7 × (−70)/3 = 𝟏𝟔𝟎/𝟑 y = 𝑧/14 = 1/14 × (−70)/3 = (−𝟓)/𝟑 Therefore, the required vector 𝑑 ⃗ = x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ = 160/3 𝑖 ̂ − 5/3 𝑗 ̂ – 70/3 𝑘 ̂ = 𝟏/𝟑 (160𝒊 ̂ − 5𝒋 ̂ – 70𝒌 ̂) Note: Answer given in the book is incorrect If we have made any mistake, please email at admin@teachoo.com

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.