Example 29 - Let a + b + c = 0, find a.b + b.c + c.a if |a| = 1 - Scalar product - Solving

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Example 29 (Method 1) Three vectors 𝑎īˇ¯, 𝑏īˇ¯ and 𝑐īˇ¯ satisfy the condition 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯ = 0īˇ¯ . Evaluate the quantity Îŧ = 𝑎īˇ¯ ⋅ 𝑏īˇ¯ + 𝑏īˇ¯ ⋅ 𝑐īˇ¯ + 𝑐īˇ¯ ⋅ 𝑎īˇ¯, if | 𝑎īˇ¯|=1, | 𝑏īˇ¯|= 4 and | cīˇ¯|= 2. Given | 𝑎īˇ¯|=1, | 𝑏īˇ¯|= 4 and | cīˇ¯|= 2 Also, 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯ = 0īˇ¯ So, 𝒂īˇ¯ + 𝒃īˇ¯ + 𝒄īˇ¯īˇ¯ = 𝟎īˇ¯īˇ¯ = 0 Now, | 𝒂īˇ¯+ 𝒃īˇ¯+ 𝒄īˇ¯ |2 = ( 𝒂īˇ¯ + 𝒃īˇ¯ + 𝒄īˇ¯) . ( 𝒂īˇ¯ + 𝒃īˇ¯ + 𝒄īˇ¯) = 𝑎īˇ¯. 𝑎īˇ¯ + 𝑎īˇ¯ . 𝑏īˇ¯ + 𝑎īˇ¯ . 𝑐īˇ¯ + 𝑏īˇ¯ . 𝑎īˇ¯ + 𝑏īˇ¯ . 𝑏īˇ¯ + 𝑏īˇ¯ . 𝑐īˇ¯ + 𝑐īˇ¯ . 𝑎īˇ¯ + 𝑐īˇ¯ . 𝑏īˇ¯ + 𝑐īˇ¯ . 𝑐īˇ¯ = 𝑎īˇ¯. 𝑎īˇ¯ + 𝑎īˇ¯ . 𝑏īˇ¯ + 𝒄īˇ¯ . 𝒂īˇ¯ + 𝒂īˇ¯ . 𝒃īˇ¯ + 𝑏īˇ¯ . 𝑏īˇ¯ + 𝑏īˇ¯ . 𝑐īˇ¯ + 𝑎īˇ¯ . 𝑐īˇ¯ + 𝒃īˇ¯ . 𝒄īˇ¯ + 𝑐īˇ¯ . 𝑐īˇ¯ = 𝑎īˇ¯ . 𝑎īˇ¯ + 𝑏īˇ¯ . 𝑏īˇ¯ + 𝑐īˇ¯ . 𝑐īˇ¯ + 2 𝑎īˇ¯. 𝑏īˇ¯ + 2 𝑏īˇ¯. 𝑐īˇ¯ + 2 𝑐īˇ¯. 𝑎īˇ¯ = 𝒂īˇ¯ . 𝒂īˇ¯ + 𝒃īˇ¯ . 𝒃īˇ¯ + 𝒄īˇ¯ . 𝒄īˇ¯ + 2( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = 𝒂īˇ¯īˇ¯đŸ + 𝒃īˇ¯īˇ¯đŸ + 𝒄īˇ¯īˇ¯đŸ + 2 ( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯ . 𝑎īˇ¯) = 12 + 42 + 22 + 2( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = 1 + 16 + 4 + 2( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = 21 + 2 ( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) So, | 𝑎īˇ¯+ 𝑏īˇ¯+ 𝑐īˇ¯ |2 = 21 + 2 ( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) Now, 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯īˇ¯ = 0 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯īˇ¯2 = 0 21 + 2 ( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = 0 2( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = −21 ( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = −21īˇŽ2īˇ¯ Therefore, 𝝁 = 𝒂īˇ¯. 𝒃īˇ¯ + 𝒃īˇ¯. 𝒄īˇ¯ + 𝒄īˇ¯ . 𝒂īˇ¯ = −𝟐𝟏īˇŽđŸīˇ¯ Example 29 (Method 2) Three vectors 𝑎īˇ¯, 𝑏īˇ¯ and 𝑐īˇ¯ satisfy the condition 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯ = 0īˇ¯ . Evaluate the quantity Îŧ = 𝑎īˇ¯ ⋅ 𝑏īˇ¯ + 𝑏īˇ¯ ⋅ 𝑐īˇ¯ + 𝑐īˇ¯ ⋅ 𝑎īˇ¯, if | 𝑎īˇ¯|=1, | 𝑏īˇ¯|= 4 and | cīˇ¯|= 2. Given | 𝑎īˇ¯|=1, | 𝑏īˇ¯|= 4 and | cīˇ¯|= 2 Also, ( 𝑎īˇ¯ + 𝑏īˇ¯ + 𝑐īˇ¯ ) = 0īˇ¯ Now, 𝒂īˇ¯ . ( 𝒂īˇ¯ + 𝒃īˇ¯ + 𝒄īˇ¯) = 𝑎īˇ¯ . 𝑎īˇ¯ + 𝑎īˇ¯. 𝑏īˇ¯ + 𝑎īˇ¯ . 𝑐īˇ¯ 𝑎īˇ¯ . 0īˇ¯ = 𝑎īˇ¯. 𝑎īˇ¯ + 𝑎īˇ¯. 𝑏īˇ¯ + 𝑎īˇ¯. 𝑐īˇ¯ 0 = 𝒂īˇ¯. 𝒂īˇ¯ + 𝑎īˇ¯. 𝑏īˇ¯ + 𝑎īˇ¯. 𝑐īˇ¯ 0īˇ¯ = 𝒂īˇ¯īˇ¯đŸ + 𝑎īˇ¯. 𝑏īˇ¯ + 𝒂īˇ¯. 𝒄īˇ¯ 0īˇ¯ = 𝑎īˇ¯īˇ¯2 + 𝑎īˇ¯. 𝑏īˇ¯ + 𝒄īˇ¯. 𝒂īˇ¯ 0 = 12 + 𝑎īˇ¯. 𝑏īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯ 0 = 1 + 𝑎īˇ¯. 𝑏īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯ 𝑎īˇ¯. 𝑏īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯ = −1 Also, 𝒃īˇ¯ . ( 𝒂īˇ¯ + 𝒃īˇ¯ + 𝒄īˇ¯) = 𝑏īˇ¯ . 𝑎īˇ¯ + 𝑏īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯ . 𝑐īˇ¯ 𝑏īˇ¯ . 0īˇ¯ = 𝑏īˇ¯. 𝑎īˇ¯ + 𝑏īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ 0 = 𝒃īˇ¯. 𝒂īˇ¯ + 𝑏īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ 0 = 𝒂īˇ¯. 𝒃īˇ¯ + 𝒃īˇ¯. 𝒃īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ 0 = 𝑎īˇ¯. 𝑏īˇ¯ + 𝒃īˇ¯īˇ¯2 + 𝑏īˇ¯ − 𝑐īˇ¯ 0 = 𝑎īˇ¯. 𝑏īˇ¯ + 42 + 𝑏īˇ¯ . 𝑐īˇ¯ 0 = 𝑎īˇ¯. 𝑏īˇ¯ + 16 + 𝑏īˇ¯ . 𝑐īˇ¯ 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ = −16 Also 𝒄īˇ¯ . ( 𝒂īˇ¯+ 𝒃īˇ¯ + 𝒄īˇ¯) = 𝑐īˇ¯ . 𝑎īˇ¯ + 𝑐īˇ¯ . 𝑏īˇ¯ + 𝑐īˇ¯ . 𝑐īˇ¯ 𝑐īˇ¯. 0īˇ¯ = 𝑐īˇ¯. 𝑎īˇ¯ + 𝑐īˇ¯. 𝑏īˇ¯ + 𝑐īˇ¯. 𝑐īˇ¯ 0 = 𝑐īˇ¯. 𝑎īˇ¯ + 𝒄īˇ¯. 𝒃īˇ¯ + 𝑐īˇ¯. 𝑐īˇ¯ 0 = 𝑐īˇ¯. 𝑎īˇ¯ + 𝒃īˇ¯. 𝒄īˇ¯ + 𝒄īˇ¯. 𝒄īˇ¯ 0 = 𝑐īˇ¯. 𝑎īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝒄īˇ¯īˇ¯2 0 = 𝑐īˇ¯. 𝑎īˇ¯ + 𝑏īˇ¯ . 𝑐īˇ¯ + 22 0 = 𝑐īˇ¯. 𝑎īˇ¯ + 𝑏īˇ¯ . 𝑐īˇ¯ + 4 𝑐īˇ¯. 𝑎īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ = −4 Adding (1), (2) and (3), ( 𝒂īˇ¯. 𝒃īˇ¯ + 𝒄īˇ¯. 𝒂īˇ¯) + ( 𝒂īˇ¯. 𝒃īˇ¯ + 𝒃īˇ¯. 𝒄īˇ¯) + ( 𝒄īˇ¯. 𝒂īˇ¯ + 𝒃īˇ¯. 𝒄īˇ¯) = −1 + (–16) + (–4) 2 𝑎īˇ¯. 𝑏īˇ¯ + 2 𝑐īˇ¯. 𝑎īˇ¯ + 2 𝑏īˇ¯. 𝑐īˇ¯ = −21 2( 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯) = −21 𝑎īˇ¯. 𝑏īˇ¯ + 𝑏īˇ¯. 𝑐īˇ¯ + 𝑐īˇ¯. 𝑎īˇ¯ = −21īˇŽ2īˇ¯ Therefore, 𝝁 = 𝒂īˇ¯. 𝒃īˇ¯ + 𝒃īˇ¯. 𝒄īˇ¯ + 𝒄īˇ¯ . 𝒂īˇ¯ = −𝟐𝟏īˇŽđŸīˇ¯

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