Ex 10.3, 11 - Show |a|b + |b|a is perpendicular to |a|b - |b|a

Ex 10.3, 11 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.3, 11 Show that |𝑎 ⃗ | 𝑏 ⃗+ |𝑏 ⃗ | 𝑎 ⃗ is perpendicular to |𝑎 ⃗ | 𝑏 ⃗− |𝑏 ⃗ | 𝑎 ⃗, for any two nonzero vectors 𝑎 ⃗ and 𝑏 ⃗ If two vectors 𝑝 ⃗ and 𝑞 ⃗ are perpendicular to each other , then their scalar (dot) product is zero, i.e. 𝒑 ⃗ . 𝒒 ⃗ = 0 Hence, to show (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) is perpendicular to (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗), We need to prove (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) . (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗) = 0 Solving L.HS. (|𝑎 ⃗ | 𝑏 ⃗ + |𝑏 ⃗ | 𝑎 ⃗) . (|𝑎 ⃗ | 𝑏 ⃗ − |𝑏 ⃗ | 𝑎 ⃗) = (|𝑎 ⃗ | 𝑏 ⃗ ) . (|𝑎 ⃗ |" " 𝑏 ⃗ ) − (|𝑎 ⃗ | 𝑏 ⃗ ) . (|𝑏 ⃗ | 𝑎 ⃗ ) + (|𝑏 ⃗ | 𝑎 ⃗ ). (|𝑎 ⃗ | 𝑏 ⃗ ) – (|𝑏 ⃗ | 𝑎 ⃗ ) . (|𝑏 ⃗ | 𝑎 ⃗ ) = |𝑎 ⃗ |2 𝑏 ⃗ . 𝑏 ⃗ − |𝑎 ⃗ ||𝑏 ⃗ | 𝒃 ⃗ . 𝒂 ⃗ + |𝑏 ⃗ | |𝑎 ⃗ | 𝑎 ⃗. 𝑏 ⃗ − |𝑏 ⃗ |2𝑎 ⃗ . 𝑎 ⃗ = |𝑎 ⃗ |2 𝑏 ⃗ . 𝑏 ⃗ − |𝑎 ⃗ ||𝑏 ⃗ | 𝒂 ⃗ . 𝒃 ⃗ + |𝑎 ⃗ ||𝑏 ⃗ | 𝑎 ⃗. 𝑏 ⃗ − |𝑏 ⃗ |2𝑎 ⃗ . 𝑎 ⃗ = |𝑎 ⃗ |2 𝒃 ⃗ . 𝒃 ⃗ − |𝑏 ⃗ |2𝒂 ⃗ . 𝒂 ⃗ = |𝑎 ⃗ |2 |𝒃 ⃗ |𝟐 −|𝑏 ⃗ |2 |𝒂 ⃗ |𝟐 = 0 = RHS Hence proved.

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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.