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Ex 10.3, 15 - If vertices of A, B, C of triangle are (1, 2, 3), (-1,0,

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Ex 10.3, 15 - Chapter 10 Class 12 Vector Algebra - Part 2

Ex 10.3, 15 - Chapter 10 Class 12 Vector Algebra - Part 3
Ex 10.3, 15 - Chapter 10 Class 12 Vector Algebra - Part 4

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Ex 10.3, 15 (Introduction) If the vertices A, B, C of a triangle ABC are (1,2,3), (–1, 0, 0), (0, 1, 2) respectively, then find ∠ABC. [∠ABC is the angle between the vectors (𝐡𝐴) βƒ— and (𝐡𝐢) βƒ—]. Consider a triangle ABC as shown ∠ ABC is not the angle between vectors (𝐴𝐡) βƒ— and (𝐡𝐢) βƒ— But the angle between vectors (𝐡𝐴) βƒ— and (𝐡𝐢) βƒ— ∴ ∠ ABC = Angle between vectors (𝐡𝐴) βƒ— and (𝐡𝐢) βƒ— Ex 10.3, 15 If the vertices A, B, C of a triangle ABC are (1,2,3), (–1, 0, 0), (0, 1, 2) respectively, then find ∠ABC. [∠ABC is the angle between the vectors (𝐡𝐴) Μ… and (𝐡𝐢) Μ…]. A (1, 2, 3) B (βˆ’1, 0, 0) C (0, 1, 2) ∠ABC = Angle b/w (𝐡𝐴) βƒ— and (𝐡𝐢) βƒ— We use formula π‘Ž βƒ—. 𝑏 βƒ— = |π‘Ž βƒ— | |𝑏 βƒ— | cos ΞΈ , ΞΈ is the angle b/w π‘Ž βƒ— & 𝑏 βƒ— We find (𝐡𝐴) βƒ— and (𝐡𝐢) βƒ— (𝐡𝐴) βƒ— = (1 βˆ’ (-1)) 𝑖 Μ‚ + (2 – 0) 𝑗 Μ‚ + (3 – 0) π‘˜ Μ‚ = 2𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚ (𝐡𝐢) βƒ— = (0 βˆ’ (βˆ’1)) 𝑖 Μ‚ + (1 βˆ’ 0) 𝑗 Μ‚ + (2 βˆ’ 0) π‘˜ Μ‚ = 1𝑖 Μ‚ + 1𝑗 Μ‚ + 2π‘˜ Μ‚ Now, (𝑩𝑨) βƒ— . (𝑩π‘ͺ) βƒ— = ("2" 𝑖 Μ‚" + " 2𝑗 Μ‚" + " 3π‘˜ Μ‚) . ("1" 𝑖 Μ‚" + " 1𝑗 Μ‚" + " 2π‘˜ Μ‚) = 2.1+2.1+3.2 = 2+2+6 = 10 Magnitude of (𝐡𝐢) βƒ— = √(12+12+22) |(𝑩π‘ͺ) βƒ— | = √(1+1+4) = βˆšπŸ” Also, (𝐡𝐴) βƒ— . (𝐡𝐢) βƒ— = |(𝐡𝐴) βƒ— | . |(𝐡𝐢) βƒ— | cos ΞΈ 10 = √17 Γ— √6 Γ— cos ΞΈ √17 Γ— √6 Γ— cos ΞΈ = 10 cos ΞΈ = 10/(√17 Γ—βˆš6) cos ΞΈ = 10/√102 ΞΈ = cosβˆ’1 (10/√102) Thus ∠ABC = cosβˆ’1 (𝟏𝟎/√𝟏𝟎𝟐).

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.