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Example 28 - Chapter 10 Class 12 Vector Algebra (Term 2)

Last updated at March 22, 2023 by

Let |a| = 3, |b| = 4, |c| = 5 and each one being perpendicular to sum

Example 28 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 28 - Chapter 10 Class 12 Vector Algebra - Part 3

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Example 28 Let π‘Ž βƒ—, 𝑏 βƒ— and 𝑐 βƒ— be three vectors such that |π‘Ž βƒ—|= 3, |𝑏 βƒ—|= 4, |𝑐 βƒ—|= 5 and each one of them being perpendicular to the sum of the other two, find |π‘Ž βƒ—" + " 𝑏 βƒ—" + " 𝑐 βƒ— |.Given, |π‘Ž βƒ—| = 3 , |𝑏 βƒ—| = 4 , |𝑐 βƒ—|= 5 Also, Each one of them being perpendicular to the sum of the other two 𝒂 βƒ— is perpendicular to 𝒃 βƒ— + 𝒄 βƒ— π‘Ž βƒ—. (𝑏 βƒ— + 𝑐 βƒ—) = 0 𝒂 βƒ—. 𝒃 βƒ— + 𝒂 βƒ—. 𝒄 βƒ— = 0 𝒃 βƒ— is perpendicular to 𝒂 βƒ— + 𝒄 βƒ— 𝑏 βƒ—. (π‘Ž βƒ— + 𝑐 βƒ—) = 0 𝒃 βƒ—. 𝒂 βƒ— + 𝒃 βƒ—. 𝒄 βƒ— = 0 𝒄 βƒ— is perpendicular to 𝒂 βƒ— + 𝒃 βƒ— 𝑐 βƒ—. (π‘Ž βƒ— + 𝑏 βƒ—) = 0 𝒄 βƒ—. 𝒂 βƒ— + 𝒄 βƒ—. 𝒃 βƒ— = 0 Now, |𝒂 βƒ—+𝒃 βƒ—+𝒄 βƒ— |2 = (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—) . (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—) = π‘Ž βƒ—. π‘Ž βƒ— + π‘Ž βƒ— . 𝑏 βƒ— + π‘Ž βƒ— . 𝑐 βƒ— + 𝑏 βƒ— . π‘Ž βƒ— + 𝑏 βƒ— . 𝑏 βƒ— + 𝑏 βƒ— . 𝑐 βƒ— + 𝑐 βƒ— . π‘Ž βƒ— + 𝑐 βƒ— . 𝑏 βƒ— + 𝑐 βƒ— . 𝑐 βƒ— = π‘Ž βƒ—. π‘Ž βƒ— + (𝒂 βƒ— . 𝒃 βƒ— + 𝒂 βƒ— . 𝒄 βƒ—) + 𝑏 βƒ— . 𝑏 βƒ— + (𝒃 βƒ— . 𝒂 βƒ— + 𝒃 βƒ— . 𝒄 βƒ—) + 𝑐 βƒ— . 𝑐 βƒ— + (𝒄 βƒ— . 𝒂 βƒ— + 𝒄 βƒ— . 𝒃 βƒ— = π‘Ž βƒ—. π‘Ž βƒ— + (0) + 𝑏 βƒ— . 𝑏 βƒ— + (0) + 𝑐 βƒ— . 𝑐 βƒ— + (0) = 𝒂 βƒ—. 𝒂 βƒ— + 𝑏 βƒ—. 𝑏 βƒ— + 𝑐 βƒ—. 𝑐 βƒ— = "|" 𝒂 βƒ—"|"2 + "|" 𝑏 βƒ—"|"2 + "|" 𝑐 βƒ—"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |𝒂 βƒ—+𝒃 βƒ—+𝒄 βƒ— |2 = 50 Taking square root both sides, "|" π‘Ž βƒ— "+ " 𝑏 βƒ—" + " 𝑐 βƒ—"|" = √50 "|" π‘Ž βƒ— "+ " 𝑏 βƒ— "+ " 𝑐 βƒ—"|" = √25 Γ— √2 "|" 𝒂 βƒ— "+ " 𝒃 βƒ— "+ " 𝒄 βƒ—"|" = πŸ“βˆšπŸ = π‘Ž βƒ—. π‘Ž βƒ— + (0) + 𝑏 βƒ— . 𝑏 βƒ— + (0) + 𝑐 βƒ— . 𝑐 βƒ— + (0) = 𝒂 βƒ—. 𝒂 βƒ— + 𝑏 βƒ—. 𝑏 βƒ— + 𝑐 βƒ—. 𝑐 βƒ— = "|" 𝒂 βƒ—"|"2 + "|" 𝑏 βƒ—"|"2 + "|" 𝑐 βƒ—"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |𝒂 βƒ—+𝒃 βƒ—+𝒄 βƒ— |2 = 50 Taking square root both sides, "|" π‘Ž βƒ— "+ " 𝑏 βƒ—" + " 𝑐 βƒ—"|" = √50 "|" π‘Ž βƒ— "+ " 𝑏 βƒ— "+ " 𝑐 βƒ—"|" = √25 Γ— √2 "|" 𝒂 βƒ— "+ " 𝒃 βƒ— "+ " 𝒄 βƒ—"|" = πŸ“βˆšπŸ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.