If a, b, c are unit vectors such that a + b + c = 0, find value of a.b

Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 3

Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 4 Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 5 Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 6 Ex 10.3, 13 - Chapter 10 Class 12 Vector Algebra - Part 7

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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 ⃗ + 2𝑎 ⃗. 𝑏 ⃗ + 2𝑏 ⃗. 𝑐 ⃗ + 2𝑐 ⃗. 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) 2𝑎 ⃗. 𝑏 ⃗ + 2𝑐 ⃗. 𝑎 ⃗ + 2𝑏 ⃗. 𝑐 ⃗ = −3 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = −3 𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗. 𝒂 ⃗ = (−𝟑)/𝟐

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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.