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Example 14 - Find angle between vectors a=i+j-k and b=i-j+k

Example 14 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 14 - Chapter 10 Class 12 Vector Algebra - Part 3

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Example 14 Find angle β€˜ΞΈβ€™ between the vectors π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚. Given π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚ 𝑏 βƒ— = 𝑖 Μ‚ – 𝑗 Μ‚ + π‘˜ Μ‚ We know that 𝒂 βƒ— . 𝒃 βƒ— = "|" 𝒂 βƒ—"|" "|" 𝒃 βƒ—"|" cos ΞΈ where ΞΈ is the angle between π‘Ž βƒ— and 𝑏 βƒ— Finding |𝒂 βƒ— |, |𝒃 βƒ— | and 𝒂 βƒ— . 𝒃 βƒ— Magnitude of π‘Ž βƒ— = √(12+1^2+(βˆ’1)2) |𝒂 βƒ— | = √(1+1+1) = βˆšπŸ‘ Magnitude of 𝑏 βƒ— = √(12+(βˆ’1)2+12) |𝒃 βƒ— | = √(1+1+1) = βˆšπŸ‘ Finding 𝒂 βƒ— . 𝒃 βƒ— 𝒂 βƒ— . 𝒃 βƒ— = (1𝑖 Μ‚ + 1𝑗 Μ‚ – 1π‘˜ Μ‚). (1𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚) = 1.1 + 1.(βˆ’1) + (βˆ’1)1 = 1 – 1 βˆ’ 1 = βˆ’1 Now, 𝒂 βƒ— . 𝒃 βƒ— = "|" 𝒂 βƒ—"|" "|" 𝒃 βƒ—"|" cos ΞΈ Putting values βˆ’1 = √3 Γ— √3 Γ— cos ΞΈ βˆ’1 = 3 cos ΞΈ cos ΞΈ = (βˆ’1)/3 ΞΈ = cosβˆ’1 ((βˆ’πŸ)/πŸ‘) Therefore, the angle between π‘Ž βƒ— and 𝑏 βƒ— is cos-1((βˆ’1)/3)

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