Question 10 (OR 2 nd question)

Find the angle between the vectors:

  a = i + j – k and b = i – j + k

Find angle between vectors: a = i + j - k and b = i - j + k - Teachoo

Question 10 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 10 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3


Transcript

Question 10 (OR 2nd question) Find the angle between the vectors: π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ – 𝑗 Μ‚ + π‘˜ Μ‚ Given π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ – 𝑗 Μ‚ + π‘˜ Μ‚ We know that π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ where ΞΈ is the angle between π‘Ž βƒ— & 𝑏 βƒ— Now, 𝒂 βƒ—. 𝒃 βƒ— = (𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚). (𝑖 Μ‚ – 𝑗 Μ‚ + π‘˜ Μ‚) = (1𝑖 Μ‚ + 1𝑗 Μ‚ – π‘˜ Μ‚). (1𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚) = 1 Γ— 1 + 1 Γ— (βˆ’1) + (-1) Γ— 1 = 1 – 1 – 1 = –1 Magnitude of π‘Ž βƒ— = √(12+12+(βˆ’1)2) |π‘Ž βƒ— |= √(1+1+1) = √3 Magnitude of 𝑏 βƒ— = √(12+(βˆ’1)2+12) |𝑏 βƒ— |= √(1+1+1) = √3 Now, π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ –1 = √3 Γ— √3 Γ— cos ΞΈ –1 = 3 cos ΞΈ cos ΞΈ = (βˆ’1)/3 ΞΈ = 𝒄𝒐𝒔^(βˆ’πŸ) ((βˆ’πŸ)/πŸ‘) Thus, the angle between π‘Ž βƒ— and 𝑏 βƒ— is cos-1((βˆ’πŸ)/πŸ‘)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.