Last updated at May 29, 2018 by Teachoo

Transcript

Misc 13 The scalar product of the vector 𝑖 + 𝑗 + 𝑘 with a unit vector along the sum of vectors 2 𝑖 + 4 𝑗 − 5 𝑘 and λ 𝑖 + 2 𝑗 + 3 𝑘 is equal to one. Find the value of λ. Let 𝑎 = 𝑖 + 𝑗 + 𝑘 = 1 𝑖 + 1 𝑗 + 1 𝑘 𝑏 = 2 𝑖 + 4 𝑗 – 5 𝑘 𝑐 = 𝜆 𝑖 + 2 𝑗 + 3 𝑘 ( 𝑏 + 𝑐) = (2 + 𝜆) 𝑖 + (4 + 2) 𝑗 + (−5 + 3) 𝑘 = (2 + 𝜆) 𝑖 + 6 𝑗 − 2 𝑘 Let 𝑟 be unit vector along ( 𝑏 + 𝑐) 𝑟 = 1𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 ( 𝑏 + 𝑐) × ( 𝑏 + 𝑐) 𝑟 = 1 2 + 𝜆2 + 62 + −22 × ((2 + 𝜆) 𝑖 + 6 𝑗 − 2 𝑘) 𝑟 = 1 22 + 𝜆2 + 4𝜆 + 36 + 4 × ((2 + 𝜆) 𝑖 + 6 𝑗 − 2 𝑘) 𝒓 = 𝟏 𝝀𝟐 + 𝟒𝝀 +𝟒𝟒 × ((2 + 𝜆) 𝒊 + 6 𝒋 − 2 𝒌) Given, 𝑎. ( 𝑟) = 1 (1 𝑖 + 1 𝑗 + 1 𝑘).( 1 𝜆2 + 4𝜆 +44 × ((2 + 𝜆) 𝑖 + 6 𝑗 − 2 𝑘)) = 1 1 𝜆2 + 4𝜆 +44 (1 𝑖 + 1 𝑗 + 1 𝑘).((𝜆 +2) 𝑖 + 6 𝑗 − 2 𝑘) = 1 (1 𝑖 + 1 𝑗 + 1 𝑘).((𝜆 +2) 𝑖 + 6 𝑗 − 2 𝑘) = 𝜆2 + 4𝜆 +44 1.(𝜆 + 2) + 1.6 + 1.(−2) = 𝜆2 + 4𝜆 +44 𝜆 + 2 + 6 − 2 = 𝜆2 + 4𝜆 +44 𝜆 + 6 = 𝝀𝟐 + 𝟒𝝀 +𝟒𝟒 Squaring both sides (𝜆 + 6)2 = 𝜆2 + 4𝜆 +442 (𝜆 + 6)2 = 𝜆2 + 4𝜆 +442 𝜆2 + 36 + 12𝜆 = 𝜆2 + 4𝜆 +44 𝜆2 – 𝜆2 + 12𝜆 – 4𝜆 = 44 – 36 8𝜆 = 8 𝜆 = 88 𝜆 = 1 So, 𝜆 = 1

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.