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Ex 10.3

Ex 10.3, 1

Ex 10.3, 2

Ex 10.3, 3 Important

Ex 10.3, 4

Ex 10.3, 5 Important

Ex 10.3, 6

Ex 10.3, 7

Ex 10.3, 8

Ex 10.3, 9 Important

Ex 10.3, 10 Important

Ex 10.3, 11 You are here

Ex 10.3, 12 Important

Ex 10.3, 13 Important

Ex 10.3, 14

Ex 10.3, 15 Important

Ex 10.3, 16 Important

Ex 10.3, 17

Ex 10.3, 18 (MCQ) Important

Last updated at March 30, 2023 by Teachoo

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.