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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

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 April 16, 2024 by Teachoo

Example 21 (Introduction) Show that the points A(−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂), B(𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) and C(7𝑖 ̂ − 𝑘 ̂) are collinear. (1) Three points collinear i.e. AB + BC = AC (2) Three position vectors collinear i.e. |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = |(𝐴𝐶) ⃗ | Example 21 Show that the points A(−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂), B(𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) and C(7𝑖 ̂ − 𝑘 ̂) are collinear. Given A (−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) B (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) C (7𝑖 ̂ + 0𝑗 ̂ − 1𝑘 ̂) 3 points A, B, C are collinear if |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = |(𝑨𝑪) ⃗ | Finding (𝑨𝑩) ⃗ , (𝑩𝑪) ⃗ , (𝑨𝑪) ⃗ (𝑨𝑩) ⃗ = (1 – (-2)) 𝑖 ̂ + (2 − 3) 𝑗 ̂ + (3 − 5) 𝑘 ̂ = 3𝒊 ̂ – 1𝒋 ̂ – 2𝒌 ̂ (𝑩𝑪) ⃗ = (7 − 1) 𝑖 ̂ + (0 − 2) 𝑗 ̂ + (-1−3) 𝑘 ̂ = 6𝒊 ̂ – 2𝒋 ̂ – 4𝒌 ̂ (𝑨𝑪) ⃗ = (7 − (-2)) 𝑖 ̂ + (0 − 3) 𝑗 ̂ + (-1 − 5) 𝑘 ̂ = 9𝒊 ̂ – 3𝒋 ̂ – 6𝒌 ̂ Magnitude of |(𝐴𝐵) ⃗ | = √(3^2+(−1)^2+(−2)^2 ) |(𝑨𝑩) ⃗ | = √(9+1+4) = √𝟏𝟒 Magnitude of |(𝐵𝐶) ⃗ | = √(6^2+(−2)^2+(−4)^2 ) |(𝑩𝑪) ⃗ | = √(36+4+16) = √56 = √(4 × 14) = 2√𝟏𝟒 Magnitude of |(𝐴𝐶) ⃗ | = √(9^2+(−3)^2+(−6)^2 ) |(𝑨𝑪) ⃗ | = √(81+9+36) = √126 = √(9 × 14) = 3√𝟏𝟒 Thus, |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = √14 + 2√14 = 3√14 = |(𝑨𝑪) ⃗ | Thus, A, B and C are collinear.