Ex 10.3

Ex 10.3, 1

Ex 10.3, 2

Ex 10.3, 3 Important

Ex 10.3, 4

Ex 10.3, 5 Important

Ex 10.3, 6

Ex 10.3, 7

Ex 10.3, 8

Ex 10.3, 9 Important

Ex 10.3, 10 Important

Ex 10.3, 11

Ex 10.3, 12 Important

Ex 10.3, 13 Important

Ex 10.3, 14

Ex 10.3, 15 Important You are here

Ex 10.3, 16 Important

Ex 10.3, 17

Ex 10.3, 18 (MCQ) Important

Last updated at April 21, 2021 by Teachoo

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 ̂ = 2𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ (𝐵𝐶) ⃗ = (0 − (−1)) 𝑖 ̂ + (1 − 0) 𝑗 ̂ + (2 − 0) 𝑘 ̂ = 1𝑖 ̂ + 1𝑗 ̂ + 2𝑘 ̂ Now, (𝑩𝑨) ⃗ . (𝑩𝑪) ⃗ = ("2" 𝑖 ̂" + " 2𝑗 ̂" + " 3𝑘 ̂) . ("1" 𝑖 ̂" + " 1𝑗 ̂" + " 2𝑘 ̂) = 2.1+2.1+3.2 = 2+2+6 = 10 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).