Last updated at May 29, 2018 by Teachoo

Transcript

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 𝐵𝐴 and 𝐵𝐶]. Consider a triangle ABC as shown ∠ ABC is not the angle between vectors 𝐴𝐵 and 𝐵𝐶 But the angle between vectors 𝐵𝐴 and 𝐵𝐶 ∴ ∠ ABC = angle between vectors 𝐵𝐴 and 𝐵𝐶 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 𝐵𝐴 and 𝐵𝐶]. A (1, 2, 3) B (−1, 0, 0) C (0, 1, 2) ∠ABC = angle b/w 𝐵𝐴 and 𝐵𝐶 We use formula 𝑎. 𝑏 = 𝑎 𝑏 cos θ , θ is the angle b/w 𝑎 & 𝑏 We find 𝐵𝐴 and 𝐵𝐶 𝐵𝐴 = (1 − (-1)) 𝑖 + (2 – 0) 𝑗 + (3 – 0) 𝑘 = 2 𝑖 + 2 𝑗 + 3 𝑘 𝐵𝐶 = (0 − (−1)) 𝑖 + (1 − 0) 𝑗 + (2 − 0) 𝑘 = 1 𝑖 + 1 𝑗 + 2 𝑘 Now, 𝑩𝑨 . 𝑩𝑪 = (2 𝑖 + 2 𝑗 + 3 𝑘) . (1 𝑖 + 1 𝑗 + 2 𝑘) = 2.1+2.1+3.2 = 2+2+6 = 10 Magnitude of 𝐵𝐴 = 22+22+32 𝑩𝑨 = 4+4+9 = 𝟏𝟕 Magnitude of 𝐵𝐶 = 12+12+22 𝑩𝑪 = 1+1+4 = 𝟔 Also, 𝐵𝐴 . 𝐵𝐶 = 𝐵𝐴 . 𝐵𝐶 cos θ 10 = 17 × 6 × cos θ 17 × 6 × cos θ = 10 cos θ = 10 17 × 6 cos θ = 10 102 θ = cos−1 10 102 Thus ∠ABC = cos−1 𝟏𝟎 𝟏𝟎𝟐.

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.