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Supplementary examples and questions from CBSE
Supplementary Example 2 Important Deleted for CBSE Board 2022 Exams
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Supplementary Exercise Q1 Deleted for CBSE Board 2022 Exams
Supplementary Exercise Q2 Deleted for CBSE Board 2022 Exams
Supplementary Exercise Q3 Deleted for CBSE Board 2022 Exams
Supplementary Exercise Q4 Important Deleted for CBSE Board 2022 Exams
Supplementary Exercise Q5 Deleted for CBSE Board 2022 Exams
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Supplementary examples and questions from CBSE
Last updated at Aug. 24, 2021 by Teachoo
Supplementary Exercise Q7 Show that the four points A, B, C and D with position vectors 4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂, −(𝑗 ̂ + 𝑘 ̂), 3𝑖 ̂ + 9𝑗 ̂ + 4𝑘 ̂ and −4𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂, respectively are co-planar 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 (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) B (−𝑗 ̂ − 𝑘 ̂) (𝑨𝑩) ⃗ = (0𝑖 ̂ − 𝑗 ̂ − 𝑘 ̂) − (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) = −4𝒊 ̂ − 6𝒋 ̂ − 2𝒌 ̂ A (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) C (3𝑖 ̂ + 9𝑗 ̂ + 4𝑘 ̂) (𝑨𝑪) ⃗ = (3𝑖 ̂ + 9𝑗 ̂ + 4𝑘 ̂) − (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) = –𝒊 ̂ + 4𝒋 ̂ + 3𝒌 ̂ A (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) D (−4𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂) (𝑨𝑫) ⃗ = (−4𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂) − (4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂) = –8𝒊 ̂ − 𝒋 ̂ + 3𝒌 ̂ [(𝐴𝐵) ⃗, (𝐴𝐶) ⃗, (𝐴𝐷) ⃗ ] = |■8(−4&−6&−2@−1&4&3@−8&−1&3)| = −4[(4×3)−(−1×3) ] − (−6) [(–1 × 3) – (–8 × 3)] + (−2)[(−1×−1)−(−8×4) ] = –4 [12+3]+6[−3+24]−2[1+32] = −4 (15) + 6 (21) − 2 (33) = −60 + 126 − 66 = −126+ 126 = 0 ∴[(𝐴𝐵) ⃗, (𝐴𝐶) ⃗, (𝐴𝐷) ⃗ ] = 0 Therefore, points A, B, C and D are coplanar.