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Supplementary Example 5 Show that the four points with position vectors 6𝑖 ̂ − 7𝑗 ̂, 16𝑖 ̂ − 19𝑗 ̂ − 4𝑘 ̂, 3𝑗 ̂ − 6𝑘 ̂ & 2𝑖 ̂ + 5𝑗 ̂ + 10𝑘 ̂ are not co-planar Let points be A = 6𝑖 ̂ – 7𝑗 ̂ B = 16𝑖 ̂ – 19𝑗 ̂ – 4𝑘 ̂ C = 3𝑗 ̂ – 6𝑘 ̂ D = 2𝑖 ̂ + 5𝑗 ̂ +10𝑘 ̂ Four points A, B, C, D are coplanar if the three vectors (𝐴𝐵) ⃗ , (𝐴𝐶) ⃗ and (𝐴𝐷) ⃗ are coplanar. i.e. [(𝑨𝑩) ⃗, (𝑨𝑪) ⃗, (𝑨𝑫) ⃗ ] = 0 A (6𝒊 ̂ – 7𝒋 ̂) B (16𝒊 ̂ – 19𝒋 ̂ – 4𝒌 ̂) (𝑨𝑩) ⃗ = (16𝑖 ̂ – 19𝑗 ̂ – 4𝑘 ̂) – (6𝑖 ̂ – 7𝑗 ̂ + 0𝑘 ̂) = (16 − 6) 𝑖 ̂ + (−19 + 7) 𝑗 ̂ – 4𝑘 ̂ = 10𝒊 ̂ − 12𝒋 ̂ − 4𝒌 ̂ A (6𝒊 ̂ – 7𝒋 ̂) C (3𝒋 ̂ – 6𝒌 ̂) (𝑨𝑪) ⃗ = (0𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) – (6𝑖 ̂ – 7𝑗 ̂ + 0𝑘 ̂) = (0 − 6) 𝑖 ̂ + (3 + 7) 𝑗 ̂ + (−6 – 0) 𝑘 ̂ = –6𝒊 ̂ + 10𝒋 ̂ – 6𝒌 ̂ A (6𝒊 ̂ – 7𝒋 ̂) D (2𝒊 ̂ + 5𝒋 ̂ +10𝒌 ̂) (𝑨𝑫) ⃗ = (2𝑖 ̂ + 5𝑗 ̂ +10𝑘 ̂) – (6𝑖 ̂ – 7𝑗 ̂ + 0𝑘 ̂) = (2 − 6) 𝑖 ̂ + (5 + 7) 𝑗 ̂ + (10 – 0) 𝑘 ̂ = –4𝒊 ̂ + 12𝒋 ̂ + 10𝒌 ̂ Now, [(𝐴𝐵) ⃗, (𝐴𝐶) ⃗, (𝐴𝐷) ⃗ ] = |■8(10&−12&−4@−6&10&−6@−4&12&10)| = 10[(10×10)−(12×−6) ] − (−12) [(−6×10)−(−4×−6)] + (−4)[(−6×12)−(−4×10) ] = 10[100+72]+12[−60−24]−4[−72+40] = 10[172]+12[−84]−4[−32] = 1720 – 1008 + 128 = 840 ∴ [(𝑨𝑩) ⃗, (𝑨𝑪) ⃗, (𝑨𝑫) ⃗ ] ≠ 0 Therefore, points A, B, C and D are not coplanar.

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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 and Computer Science at Teachoo