Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams You are here

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 29 (Supplementary NCERT) Show that the four points A, B, C and D with position vectors 4𝑖 ̂ + 5𝑗 ̂ + 𝑘 ̂, −(𝑗 ̂ + 𝑘 ̂), 3𝑖 ̂ + 9𝑗 ̂ + 4𝑘 ̂ & −4𝑖 ̂ + 4𝑗 ̂ + 4𝑘 ̂, respectively coplanar 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.