Example 17 - Chapter 10 Class 12 Vector Algebra

Last updated at August 19, 2026 by

Example 17 - Find |a - b|, if |a| = 2|b| = 3 and a.b = 4

Example 17 - Chapter 10 Class 12 Vector Algebra - Part 2

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Example 17 Find |π‘Ž βƒ— βˆ’ 𝑏 βƒ—| , if two vectors a and b are such that |π‘Ž βƒ—| = 2, |𝑏 βƒ—| = 3 and π‘Ž βƒ— β‹… 𝑏 βƒ— = 4. Given, |𝒂 βƒ— | = 2 , |𝒃 βƒ— | = 2 & 𝒂 βƒ—. 𝒃 βƒ— = 4 We need to find |π‘Ž βƒ—βˆ’π‘ βƒ— | Taking square, |𝒂 βƒ—βˆ’π’ƒ βƒ— |2 = (𝒂 βƒ—βˆ’π’ƒ βƒ—). (𝒂 βƒ—βˆ’π’ƒ βƒ—) = π‘Ž βƒ— . π‘Ž βƒ— βˆ’ π‘Ž βƒ— . 𝑏 βƒ— βˆ’ 𝒃 βƒ—. 𝒂 βƒ— + 𝑏 βƒ— . 𝑏 βƒ— = π‘Ž βƒ— . π‘Ž βƒ— βˆ’ π‘Ž βƒ— . 𝑏 βƒ— βˆ’ 𝒂 βƒ— . 𝒃 βƒ— + 𝑏 βƒ— . 𝑏 βƒ— = 𝒂 βƒ—. 𝒂 βƒ— βˆ’ 2π‘Ž βƒ— . 𝑏 βƒ— + 𝒃 βƒ—. 𝒃 βƒ— = |𝒂 βƒ— |2 βˆ’ 2π‘Ž βƒ—. 𝑏 βƒ— + |𝒃 βƒ— |2 = 22 – 2 Γ— 4 + 32 = 4 βˆ’ 8 + 9 = 5 Thus, |𝒂 βƒ—βˆ’π’ƒ βƒ— |2 = 5 |π‘Ž βƒ—βˆ’π‘ βƒ— | = Β± √5 Since magnitude is not negative, So, |𝒂 βƒ—βˆ’π’ƒ βƒ— | = βˆšπŸ“

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