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Ex 10.4, 7 - Show that a x (b + c) =  a x b + a x c - Ex 10.4

Ex 10.4, 7 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.4, 7 - Chapter 10 Class 12 Vector Algebra - Part 3

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Ex 10.4, 7 Let the vectors π‘Ž βƒ— 𝑏 βƒ—, 𝑐 βƒ— be given as π‘Ž_1 𝑖 Μ‚ + π‘Ž_2 𝑗 Μ‚ +π‘Ž_3 π‘˜ Μ‚, 𝑏_1 𝑖 Μ‚ + 𝑏_2 𝑗 Μ‚ +𝑏_3 π‘˜ Μ‚, 𝑐_1 𝑖 Μ‚ + 𝑐_2 𝑗 Μ‚ +𝑐_3 π‘˜ Μ‚ Then show that π‘Ž βƒ— Γ— (𝑏 βƒ— + 𝑐 βƒ—) =π‘Ž βƒ— ×𝑏 βƒ— + π‘Ž βƒ— Γ— 𝑐 βƒ—. Let π‘Ž βƒ— = π‘Ž_1 𝑖 Μ‚ + π‘Ž_2 𝑗 Μ‚ +π‘Ž_3 π‘˜ Μ‚ 𝑏 βƒ— = 𝑏_1 𝑖 Μ‚ + 𝑏_2 𝑗 Μ‚ +𝑏_3 π‘˜ Μ‚, 𝑐 βƒ— = 𝑐_1 𝑖 Μ‚ + 𝑐_2 𝑗 Μ‚ + 𝑐_3 π‘˜ Μ‚ We need to show : π‘Ž βƒ— Γ— (𝑏 βƒ— + 𝑐 βƒ—) = π‘Ž βƒ— Γ— 𝑏 βƒ— + π‘Ž βƒ— Γ— 𝑐 βƒ— RHS (π‘Ž βƒ— Γ— 𝑏 βƒ—) = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@π‘Ž1&π‘Ž2&π‘Ž3@𝑏1&𝑏2&𝑏3)| = (π‘Ž2 𝑏3 βˆ’ 𝑏2 π‘Ž3) 𝑖 Μ‚ βˆ’ (π‘Ž1 𝑏3 βˆ’ 𝑏1 π‘Ž3) 𝑗 Μ‚ + (π‘Ž1 𝑏2 βˆ’ 𝑏1 π‘Ž2) π‘˜ Μ‚ (π‘Ž βƒ— Γ— 𝑐 βƒ—) = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@π‘Ž1&π‘Ž2&π‘Ž3@𝑐1&𝑐2&𝑐3)| = (π‘Ž2" " 𝑐3 βˆ’ 𝑐2" " π‘Ž3) 𝑖 Μ‚ βˆ’ (π‘Ž1" " 𝑐3 βˆ’ 𝑐1" " π‘Ž3) 𝑗 Μ‚ + (π‘Ž1" " 𝑐2 βˆ’ 𝑐1" " π‘Ž2) π‘˜ Μ‚ (𝒂 βƒ— Γ— 𝒃 βƒ—) + (𝒂 βƒ— Γ— 𝒄 βƒ—) = [π‘Ž2" " 𝑏3βˆ’π‘2 π‘Ž3+π‘Ž2 𝑐3βˆ’π‘2π‘Ž3] 𝑖 Μ‚ βˆ’ [π‘Ž1" " 𝑏3βˆ’π‘1 π‘Ž3+π‘Ž1 𝑐3βˆ’π‘1π‘Ž3] 𝑗 Μ‚ + [π‘Ž1" " 𝑏2βˆ’π‘1 π‘Ž2+π‘Ž1 𝑐2βˆ’π‘1π‘Ž2] π‘˜ Μ‚ Since the corresponding components are equal, So, π‘Ž βƒ— Γ— (𝑏 βƒ— + 𝑐 βƒ—) = π‘Ž βƒ— Γ— 𝑏 βƒ— + π‘Ž βƒ— Γ— 𝑐 βƒ— Hence proved. Ex 10.4, 7 Let the vectors , be given as 1 + 2 + 3 , 1 + 2 + 3 , 1 + 2 + 3 Then show that ( + ) = + . Let = 1 + 2 + 3 = 1 + 2 + 3 , = 1 + 2 + 3 We need to show : ( + ) = + LHS ( + ) = ( 1 + 1) + ( 2 + 2) + ( 3 + 3) ( + ) = 1 ( 1+ 1) 2 ( 2+ 2) 3 ( 3+ 3) = 2 ( 3 + 3) ( 2 + 2) 3 1 ( 3 + 3) ( 1 + 1) 3 + 1 ( 2 + 2) ( 1 + 1) 2 = 2 3+ 2 3 2 3 2 3 1 3 1 3+ 1 3 1 3 + 1 2+ 1 2 1 2 1 2

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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.