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Supplementary Exercise Q6

Last updated at May 29, 2023 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 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.