Supplementary Exercise Q6

Last updated at Dec. 16, 2024 by

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Supplementary Exercise Q6 if 𝑖 ̂, 𝑗 ̂ and 𝑘 ̂ are three mutually perpendicular vectors, prove that 𝑖 ̂.(𝑘 ̂×𝑗 ̂ ) = 𝑗 ̂.(𝑖 ̂×𝑘 ̂ ) = 𝑘 ̂.(𝑗 ̂ × 𝑖 ̂) = −1 Here, 𝑖 ̂ = 1𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂ 𝑗 ̂ = 0𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂ 𝑘 ̂ = 0𝑖 ̂ + 0𝑗 ̂ + 1𝑘 ̂ 𝒊 ̂.(𝒌 ̂×𝒋 ̂ ) = [■8(𝑖 ̂&𝑘 ̂&𝑗 ̂ )] = [■8(1&0&0@0&0&1@0&1&0)] = 1(0 – 1) – 0(0 – 0) + 0(0 – 0) = –1 𝒋 ̂.(𝒊 ̂×𝒌 ̂ ) = [■8(𝑗 ̂&𝑖 ̂&𝑘)] = [■8(0&1&0@1&0&0@0&0&1)] = 0(0 – 0) – 1(1 – 0) + 0(0 – 0) = –1 𝒌 ̂.(𝒋 ̂×𝒊 ̂ ) = [■8(𝑘 ̂&𝑗 ̂&𝑖 ̂ )] = [■8(0&0&1@0&1&0@1&0&0)] = 0(0 – 0) – 0(0 – 0) + 1(0 – 1) = –1

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo