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

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Transcript

Ex 10.4, 12 Area of a rectangle having vertices A, B, C & D with position vectors −𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂, 𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂, 𝑖 ̂ − 1/2 𝑗 ̂ + 4𝑘 ̂, −𝑖 ̂ − 1/2 𝑗 ̂ + 4𝑘 ̂ respectively is (A) 1/2 (B) 1 (C) 2 (D) 4 (𝑂𝐴) ⃗ = −𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂ = −1𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂ (𝑂𝐵) ⃗ = 𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂ = 1𝑖 ̂ + 1/2 𝑗 ̂ + 4𝑘 ̂ (𝑂𝐶) ⃗ = 𝑖 ̂ − 1/2 𝑗 ̂ + 4𝑘 ̂ = 1𝑖 ̂ - 1/2 𝑗 ̂ + 4𝑘 ̂ (𝑂𝐷) ⃗ = −𝑖 ̂ − 1/2 𝑗 ̂ + 4𝑘 ̂ = −1𝑖 ̂ – 1/2 𝑗 ̂ + 4𝑘 ̂ Since rectangle is also a parallelogram Area of rectangle ABCD = |(𝐴𝐵) ⃗×(𝐵𝐶) ⃗ | (𝑨𝑩) ⃗ = (𝑶𝑩) ⃗ − (𝑶𝑨) ⃗ = ("1" 𝑖 ̂" + " 1/2 " " 𝑗 ̂" + 4" 𝑘 ̂ ) − ("−1" 𝑖 ̂" + " 1/2 " " 𝑗 ̂" + 4" 𝑘 ̂ ) = ("1−(−1)" 𝑖 ̂" +" (1/2−1/2) 𝑗 ̂" + (4 − 4)" ) 𝑘 ̂ = 2𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂ (𝑩𝑪) ⃗ = (𝑶𝑪) ⃗ − (𝑶𝑩) ⃗ = ("1" 𝑖 ̂" − " 1/2 " " 𝑗 ̂" + 4" 𝑘 ̂ ) − ("1" 𝑖 ̂" + " 1/2 " " 𝑗 ̂" + 4" 𝑘 ̂ ) = ("(1−1)" 𝑖 ̂" +" (−1/2−1/2) 𝑗 ̂" + (4 − 4)" ) 𝑘 ̂ = 0𝑖 ̂ – 1𝑗 ̂ + 0𝑘 ̂ |(𝑨𝑩) ⃗×(𝑩𝑪) ⃗ | = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&0&0@0&−1&0)| = 𝑖 ̂ (0 × 0 − (−1) × 0) − 𝑗 ̂ (2 × 0 − 0 × 0) + 𝑘 ̂ (2 × −1 − 0 × 0 ) = 𝑖 ̂ (0 − 0) − 𝑗 ̂ (0 − 0) + 𝑘 ̂ (−2 − 0) = 0𝒊 ̂ − 0𝒋 ̂ − 2𝒌 ̂ Now, Magnitude of (𝐴𝐵) ⃗×(𝐵𝐶) ⃗ = √(02+02+(−2)2) |(𝑨𝑩) ⃗×(𝑩𝑪) ⃗ | = √4 = 2 Therefore, area of rectangle ABCD = |(𝐴𝐵) ⃗×(𝐵𝐶) ⃗ | = 2 Hence, (C) is the correct option

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.