Ex 10.4, 12 - Area of a rectangle having vertices A, B, C, D

Ex 10.4, 12 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.4, 12 - Chapter 10 Class 12 Vector Algebra - Part 3

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

Transcript

Ex 10.4, 12 Area of a rectangle having vertices A, B, C & D with position vectors βˆ’π‘– Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚, 𝑖 Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚, 𝑖 Μ‚ βˆ’ 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚, βˆ’π‘– Μ‚ βˆ’ 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ respectively is (A) 1/2 (B) 1 (C) 2 (D) 4 (𝑂𝐴) βƒ— = βˆ’π‘– Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ (𝑂𝐡) βƒ— = 𝑖 Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ = 1𝑖 Μ‚ + 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ (𝑂𝐢) βƒ— = 𝑖 Μ‚ βˆ’ 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ = 1𝑖 Μ‚ - 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ (𝑂𝐷) βƒ— = βˆ’π‘– Μ‚ βˆ’ 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ – 1/2 𝑗 Μ‚ + 4π‘˜ Μ‚ Since rectangle is also a parallelogram Area of rectangle ABCD = |(𝐴𝐡) βƒ—Γ—(𝐡𝐢) βƒ— | (𝑨𝑩) βƒ— = (𝑢𝑩) βƒ— βˆ’ (𝑢𝑨) βƒ— = ("1" 𝑖 Μ‚" + " 1/2 " " 𝑗 Μ‚" + 4" π‘˜ Μ‚ ) βˆ’ ("βˆ’1" 𝑖 Μ‚" + " 1/2 " " 𝑗 Μ‚" + 4" π‘˜ Μ‚ ) = ("1βˆ’(βˆ’1)" 𝑖 Μ‚" +" (1/2βˆ’1/2) 𝑗 Μ‚" + (4 βˆ’ 4)" ) π‘˜ Μ‚ = 2𝑖 Μ‚ + 0𝑗 Μ‚ + 0π‘˜ Μ‚ (𝑩π‘ͺ) βƒ— = (𝑢π‘ͺ) βƒ— βˆ’ (𝑢𝑩) βƒ— = ("1" 𝑖 Μ‚" βˆ’ " 1/2 " " 𝑗 Μ‚" + 4" π‘˜ Μ‚ ) βˆ’ ("1" 𝑖 Μ‚" + " 1/2 " " 𝑗 Μ‚" + 4" π‘˜ Μ‚ ) = ("(1βˆ’1)" 𝑖 Μ‚" +" (βˆ’1/2βˆ’1/2) 𝑗 Μ‚" + (4 βˆ’ 4)" ) π‘˜ Μ‚ = 0𝑖 Μ‚ – 1𝑗 Μ‚ + 0π‘˜ Μ‚ |(𝑨𝑩) βƒ—Γ—(𝑩π‘ͺ) βƒ— | = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@2&0&0@0&βˆ’1&0)| = 𝑖 Μ‚ (0 Γ— 0 βˆ’ (βˆ’1) Γ— 0) βˆ’ 𝑗 Μ‚ (2 Γ— 0 βˆ’ 0 Γ— 0) + π‘˜ Μ‚ (2 Γ— βˆ’1 βˆ’ 0 Γ— 0 ) = 𝑖 Μ‚ (0 βˆ’ 0) βˆ’ 𝑗 Μ‚ (0 βˆ’ 0) + π‘˜ Μ‚ (βˆ’2 βˆ’ 0) = 0π’Š Μ‚ βˆ’ 0𝒋 Μ‚ βˆ’ 2π’Œ Μ‚ Now, Magnitude of (𝐴𝐡) βƒ—Γ—(𝐡𝐢) βƒ— = √(02+02+(βˆ’2)2) |(𝑨𝑩) βƒ—Γ—(𝑩π‘ͺ) βƒ— | = √4 = 2 Therefore, area of rectangle ABCD = |(𝐴𝐡) βƒ—Γ—(𝐡𝐢) βƒ— | = 2 Hence, (C) is the correct option

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.