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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Example 27 If 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚, 2𝑖 Μ‚ + 5𝑗 Μ‚, 3𝑖 Μ‚ + 2𝑗 Μ‚ – 3π‘˜ Μ‚ and 𝑖 Μ‚ – 6𝑗 Μ‚ – π‘˜ Μ‚ are the position vectors of points A, B, C and D respectively, then find the angle between (𝐴𝐡) βƒ— and (𝐢𝐷) βƒ— . Deduce that (𝐴𝐡) βƒ— and (𝐢𝐷) βƒ— are collinear.Angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is given by cos ΞΈ = ((𝑨𝑩) βƒ— . (π‘ͺ𝑫) βƒ—)/|(𝑨𝑩) βƒ— ||(π‘ͺ𝑫) βƒ— | A(π’Š Μ‚ + 𝒋 Μ‚ + π’Œ Μ‚), B(2π’Š Μ‚ + 5𝒋 Μ‚) (𝐴𝐡) βƒ— = (2 βˆ’ 1) 𝑖 Μ‚ + (5 βˆ’ 1) 𝑗 Μ‚ + (0 βˆ’ 1) π‘˜ Μ‚ = 1𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ π‘˜ Μ‚ |(𝐴𝐡) βƒ— | = √(1^2+4^2+(βˆ’1)^2 ) = √18 = √(9 Γ— 2) = 3√2 C(3π’Š Μ‚ + 2𝒋 Μ‚ – 3π’Œ Μ‚), D(π’Š Μ‚ – 6𝒋 Μ‚ – π’Œ Μ‚) (𝐢𝐷) βƒ— = (1 βˆ’ 3) 𝑖 Μ‚ + (–6 βˆ’ 2) 𝑗 Μ‚ + (βˆ’1 βˆ’ (-3)) π‘˜ Μ‚ = –2𝑖 Μ‚ – 8𝑗 Μ‚ + 2π‘˜ Μ‚ |(𝐢𝐷) βƒ— | = √((βˆ’2)^2+(βˆ’8)^2+2^2 ) = √72 = √(36 Γ— 2) = 6√2 Now, cos ΞΈ = ((𝐴𝐡) βƒ— . (𝐢𝐷) βƒ—)/|(𝐴𝐡) βƒ— ||(𝐢𝐷) βƒ— | = ((𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ π‘˜ Μ‚ ).(βˆ’2𝑖 Μ‚ βˆ’ 8𝑗 Μ‚ + 2π‘˜ Μ‚ ))/(3√2 Γ— 6√2) = (1(βˆ’2) + 4(βˆ’8) βˆ’ 1(2))/(3√2 Γ— 6√2) = (βˆ’2 βˆ’ 32 βˆ’ 2)/(3 Γ— 6 Γ— √2 Γ— √2) = (βˆ’36)/36 = –1 Since cos ΞΈ = –1, ΞΈ = 180Β° So, ΞΈ = 180Β° = 180Β° Γ— πœ‹/180 = Ο€ So, angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is Ο€ Also, Since angle between (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— is 180Β° , they are in opposite directions Since (𝐴𝐡) βƒ— & (𝐢𝐷) βƒ— are parallel to the same line π‘š βƒ—, they are collinear.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.