Find the Projection (vector) of 2i − j + k on i − 2j + k

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Question 11 Find the Projection (vector) of 2๐‘–ย ฬ‚ โˆ’ ๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚ on ๐‘–ย ฬ‚ โˆ’ 2๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚ Let a = 2๐‘–ย ฬ‚ โˆ’ ๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚ and b = ๐‘–ย ฬ‚ โˆ’ 2๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚ We need to find Projection (vector) of ๐‘Žย โƒ— on ๐‘ย โƒ— Theory We know that Projection of ๐‘Žย โƒ— on ๐‘ย โƒ— = 1/("|" ๐‘ย โƒ—"|" ) (๐‘Žย โƒ—. ๐‘ย โƒ—) But here, we are asked projection (vector) So, we multiply projection by ๐‘ย โƒ—/|๐‘ย โƒ— | Projection (vector) of ๐‘Žย โƒ— on ๐‘ย โƒ— = 1/|๐‘ย โƒ— | (๐‘Žย โƒ—. ๐‘ย โƒ—) ร— ๐‘ย โƒ—/|๐‘ย โƒ— | = ((๐‘Žย โƒ— . ๐‘ย โƒ— ))/|๐‘ย โƒ— |^2 ร— ๐‘ย โƒ— (๐’‚ย โƒ—. ๐’ƒย โƒ—) = (2 ร— 1) + (โ€“1 ร— โ€“2) + (1 ร— 1) = 2 + 2 + 1 = 5 Magnitude of ๐‘ย โƒ— = โˆš(12+(โˆ’2)2+12) |๐‘ย โƒ— | = โˆš(1+4+1) = โˆš6 Projection (vector) of ๐’‚ย โƒ— on ๐’ƒย โƒ— = ((๐‘Žย โƒ— . ๐‘ย โƒ— ))/|๐‘ย โƒ— |^2 ร— ๐‘ย โƒ— = 5/(โˆš6)^2 ร— ( ๐‘–ย ฬ‚ โˆ’ 2๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚) = ๐Ÿ“/๐Ÿ” ( ๐’Šย ฬ‚ โˆ’ 2๐’‹ย ฬ‚ + ๐’Œย ฬ‚)

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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.