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

Examples

Example 1
Important

Example 2 Important

Example 3 Important

Example 4

Example 5

Example 6 Important

Example 7

Example 8 Important

Example 9

Example 10

Example 11 (i) Important

Example 11 (ii) Important

Example 12 Important

Example 13

Example 14 Important

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20

Example 21 Important You are here

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important

Example 29 Important

Example 30 Important

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 May 29, 2023 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.