Ex 10.3, 2 - Chapter 10 Class 12 Vector Algebra (Important Question)

Last updated at April 16, 2024 by Teachoo

Ex 10.3, 2 - Find angle between vectors i - 2j + 3k, 3i - 2j + k

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

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

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Transcript

Ex 10.3, 2 Find the angle between the vectors 𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂ and 3𝑖 ̂ - 2𝑗 ̂ + 𝑘 ̂Let 𝑎 ⃗ = 𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂ = 1𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂ and 𝑏 ⃗ = 3𝑖 ̂ − 2𝑗 ̂ + 𝑘 ̂ = 3𝑖 ̂ − 2𝑗 ̂ + 1𝑘 ̂ We know that 𝑎 ⃗ . 𝑏 ⃗ = |𝑎 ⃗ ||𝑏 ⃗ | cos θ ; θ is the angle between 𝑎 ⃗ & 𝑏 ⃗ Now, 𝒂 ⃗. 𝒃 ⃗ = (1𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂). (3𝑖 ̂ − 2𝑗 ̂ + 1𝑘 ̂) = 1.3 + (−2).(−2) + 3.1 = 3 + 4 + 3 = 10 Magnitude of 𝑎 ⃗ = √(12+(−2)2+32) |𝑎 ⃗ |= √(1+4+9) = √14 Magnitude of 𝑏 ⃗ = √(32+(−2)2+12) |𝑏 ⃗ |= √(9+4+1) = √14 Now, 𝑎 ⃗ . 𝑏 ⃗ = |𝑎 ⃗ ||𝑏 ⃗ | cos θ 10 = √14 × √14 x cos θ 10 = 14 × cos θ cos θ = 10/14 θ = cos-1(𝟓/𝟕) Thus, the angle between 𝑎 ⃗ and 𝑏 ⃗ is cos-1(5/7)

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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

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