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Supplementary examples and questions from CBSE
Supplementary Example 2 Important Deleted for CBSE Board 2023 Exams
Supplementary Example 3 Deleted for CBSE Board 2023 Exams
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Supplementary Exercise Q1 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q2 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q3 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q4 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q5 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q6 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q7 Deleted for CBSE Board 2023 Exams
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Supplementary Exercise Q10 Deleted for CBSE Board 2023 Exams
Supplementary examples and questions from CBSE
Last updated at March 30, 2023 by Teachoo
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&−[email protected]−6&10&−[email protected]−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.