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Ex 10.3, 13 - If a + b + c = 0, find value of a.b + b.c + c.a - Ex 10.3

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  1. Class 12
  2. Important Question for exams Class 12
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Ex 10.3, 13 (Method 1) If 𝑎﷯ , 𝑏﷯, 𝑐﷯ are unit vectors such that 𝑎﷯ + 𝑏﷯ + 𝑐﷯ = 0﷯, find the value of 𝑎﷯ . 𝑏﷯+ 𝑏﷯ . 𝑐﷯ + 𝑐﷯. 𝑎﷯ . Given 𝑎﷯ , 𝑏﷯, 𝑐﷯ are unit vectors Magnitude of 𝑎﷯ , 𝑏﷯, 𝑐﷯ is 1 So, 𝒂﷯﷯ = 𝒃﷯﷯ = 𝒄﷯﷯ = 1 Also, 𝑎﷯ + 𝑏﷯ + 𝑐﷯ = 0﷯ So, 𝒂﷯ + 𝒃﷯ + 𝒄﷯﷯ = 𝟎﷯﷯ = 0 Now, | 𝒂﷯+ 𝒃﷯+ 𝒄﷯ |2 = ( 𝒂﷯ + 𝒃﷯ + 𝒄﷯) . ( 𝒂﷯ + 𝒃﷯ + 𝒄﷯) = 𝑎﷯. 𝑎﷯ + 𝑎﷯ . 𝑏﷯ + 𝒂﷯ . 𝒄﷯ + 𝒃﷯ . 𝒂﷯ + 𝑏﷯ . 𝑏﷯ + 𝑏﷯ . 𝑐﷯ + 𝑐﷯ . 𝑎﷯ + 𝒄﷯ . 𝒃﷯ + 𝑐﷯ . 𝑐﷯ = 𝑎﷯. 𝑎﷯ + 𝑎﷯ . 𝑏﷯ + 𝒄﷯ . 𝒂﷯ + 𝒂﷯ . 𝒃﷯ + 𝑏﷯ . 𝑏﷯ + 𝑏﷯ . 𝑐﷯ + 𝑐﷯ . 𝑎﷯ + 𝒃﷯ . 𝒄﷯ + 𝑐﷯ . 𝑐﷯ = 𝑎﷯ . 𝑎﷯ + 𝑏﷯ . 𝑏﷯ + 𝑐﷯ . 𝑐﷯ + 2 𝑎﷯. 𝑏﷯ + 2 𝑏﷯. 𝑐﷯ + 2 𝑐﷯. 𝑎﷯ = 𝒂﷯ . 𝒂﷯ + 𝒃﷯ . 𝒃﷯ + 𝒄﷯ . 𝒄﷯ + 2( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = 𝒂﷯﷯𝟐 + 𝒃﷯﷯𝟐 + 𝒄﷯﷯𝟐 + 2 ( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯ . 𝑎﷯) = 12 + 12 + 12 + 2( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = 1 + 1 + 1 + 2( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = 3 + 2 ( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) ∴ | 𝑎﷯+ 𝑏﷯+ 𝑐﷯ |2 = 3 + 2 ( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) Now, 𝑎﷯ + 𝑏﷯ + 𝑐﷯﷯ = 0 𝑎﷯ + 𝑏﷯ + 𝑐﷯﷯﷮2﷯ = 0 3 + 2 ( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = 0 2( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = −3 ( 𝒂﷯. 𝒃﷯ + 𝒃﷯. 𝒄﷯ + 𝒄﷯. 𝒂﷯) = −𝟑﷮𝟐﷯ Ex 10.3, 13 (Method 2) If 𝑎﷯ , 𝑏﷯, 𝑐﷯ are unit vectors such that 𝑎﷯ + 𝑏﷯ + 𝑐﷯ = 0, find the value of 𝑎﷯ . 𝑏﷯+ 𝑏﷯ . 𝑐﷯ + 𝑐﷯. 𝑎﷯ . Given 𝑎﷯ , 𝑏﷯, 𝑐﷯ are unit vectors So, 𝒂﷯﷯ = 𝒃﷯﷯ = 𝒄﷯﷯ = 1 Also, ( 𝑎﷯ + 𝑏﷯ + 𝑐﷯ ) = 0﷯ Now, 𝒂﷯ . ( 𝒂﷯ + 𝒃﷯ + 𝒄﷯) = 𝑎﷯ . 𝑎﷯ + 𝑎﷯. 𝑏﷯ + 𝑎﷯ . 𝑐﷯ 𝑎﷯ . 0﷯ = 𝑎﷯. 𝑎﷯ + 𝑎﷯. 𝑏﷯ + 𝑎﷯. 𝑐﷯ 0 = 𝒂﷯. 𝒂﷯ + 𝑎﷯. 𝑏﷯ + 𝑎﷯. 𝑐﷯ 0﷯ = 𝒂﷯﷯𝟐 + 𝑎﷯. 𝑏﷯ + 𝑎﷯. 𝑐﷯ 0﷯ = 𝑎﷯﷯2 + 𝑎﷯. 𝑏﷯ + 𝑎﷯. 𝑐﷯ 0﷯ = 𝑎﷯﷯2 + 𝑎﷯. 𝑏﷯ + 𝒄﷯. 𝒂﷯ 0 = 12 + 𝑎﷯. 𝑏﷯ + 𝑐﷯. 𝑎﷯ 𝑎﷯. 𝑏﷯ + 𝑐﷯. 𝑎﷯ = −1 Also, 𝒃﷯ . ( 𝒂﷯ + 𝒃﷯ + 𝒄﷯) = 𝑏﷯ . 𝑎﷯ + 𝑏﷯. 𝑏﷯ + 𝑏﷯ . 𝑐﷯ 𝑏﷯ . 0﷯ = 𝑏﷯. 𝑎﷯ + 𝑏﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ 0 = 𝒃﷯. 𝒂﷯ + 𝑏﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ 0 = 𝒂﷯. 𝒃﷯ + 𝑏﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ 0 = 𝑎﷯. 𝑏﷯ + 𝒃﷯. 𝒃﷯ + 𝑏﷯. 𝑐﷯ 0 = 𝑎﷯. 𝑏﷯ + 𝒃﷯﷯2 + 𝑏﷯ . 𝑐﷯ 0 = 𝑎﷯. 𝑏﷯ + 12 + 𝑏﷯ . 𝑐﷯ 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯ = −1 Also 𝒄﷯ . ( 𝒂﷯+ 𝒃﷯ + 𝒄﷯) = 𝑐﷯ . 𝑎﷯ + 𝑐﷯ . 𝑏﷯ + 𝑐﷯ . 𝑐﷯ 𝑐﷯. 0﷯ = 𝑐﷯. 𝑎﷯ + 𝑐﷯. 𝑏﷯ + 𝑐﷯. 𝑐﷯ 0 = 𝑐﷯. 𝑎﷯ + 𝒄﷯. 𝒃﷯ + 𝑐﷯. 𝑐﷯ 0 = 𝑐﷯. 𝑎﷯ + 𝒃﷯. 𝒄﷯ + 𝑐﷯. 𝑐﷯ 0 = 𝑐﷯. 𝑎﷯ + 𝑏﷯. 𝑐﷯ + 𝒄﷯. 𝒄﷯ 0 = 𝑐﷯. 𝑎﷯ + 𝑏﷯. 𝑐﷯ + 𝑐﷯﷯2 0 = 𝑐﷯. 𝑎﷯+ 𝑏﷯ . 𝑐﷯ + 12 𝑐﷯. 𝑎﷯ + 𝑏﷯. 𝑐﷯ = −1 Adding (1), (2) and (3), ( 𝑎﷯. 𝑏﷯ + 𝑐﷯. 𝑎﷯) + ( 𝑎﷯. 𝑏﷯ + 𝑏﷯. 𝑐﷯) + ( 𝑐﷯. 𝑎﷯ + 𝑏﷯. 𝑐﷯) = −1 + (–1) + (–1) 2 𝑎﷯. 𝑏﷯ + 2 𝑐﷯. 𝑎﷯ + 2 𝑏﷯. 𝑐﷯ = −3 2( 𝑎﷯. 𝑏﷯ + 𝑏. 𝑐﷯ + 𝑐﷯. 𝑎﷯) = −3 𝒂﷯. 𝒃﷯ + 𝒃﷯. 𝒄﷯ + 𝒄﷯. 𝒂﷯ = −𝟑﷮𝟐﷯

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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