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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo