Question 22

If vector p = i + j + k and q = i – 2j + k, find a vector of magnitude 5√3 units perpendicular to the vector q and coplanar with vectors p and q

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  1. Class 12
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Transcript

Question 22 If ๐‘ โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚ and ๐‘ž โƒ— = ๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚, find a vector of magnitude 5โˆš3 units perpendicular to the vector ๐‘ž โƒ— and coplanar with vectors ๐‘ โƒ— and ๐‘ž โƒ— Let ๐‘Ÿ โƒ— be the required vector ๐‘Ÿ โƒ— = a๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚ Now, given that ๐‘Ÿ โƒ— & ๐‘ž โƒ— are perpendicular Therefore, ๐‘Ÿ โƒ—.๐‘ž โƒ— = 0 (a๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚) . (๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 0 a ร— 1 + b ร— (โ€“2) + c ร— 1 = 0 a โ€“ 2b + c = 0 Also, given that ๐‘Ÿ โƒ— is coplanar with vectors ๐‘ โƒ— and ๐‘ž โƒ— Therefore, [๐’‘ โƒ—" " ๐’’ โƒ—" " ๐’“ โƒ— ] = 0 Finding [๐’‘ โƒ—" " ๐’’ โƒ—" " ๐’“ โƒ— ] [๐‘ โƒ—" " ๐‘ž โƒ—" " ๐‘Ÿ โƒ— ] = |โ– 8(1&1&1@1&โˆ’2&1@๐‘Ž&๐‘&๐‘)| = 1[โˆ’2๐‘โˆ’๐‘ ] โˆ’ 1[๐‘โˆ’๐‘Ž] + 1[๐‘+2๐‘Ž] = โˆ’2๐‘โˆ’๐‘โˆ’๐‘+๐‘Ž+๐‘+2๐‘Ž = โˆ’3๐‘+3๐‘Ž Since [๐‘ โƒ—" " ๐‘ž โƒ—" " ๐‘Ÿ โƒ— ] = 0 โˆ’3๐‘+3๐‘Ž=0 3๐‘Ž=3๐‘ ๐‘Ž=๐‘ Now, from (1) a โ€“ 2b + c = 0 Putting c = a a โ€“ 2b + a = 0 2a โ€“ 2b = 0 2a = 2b a = b Therefore, a = b = c So, our vector ๐‘Ÿ โƒ— becomes ๐‘Ÿ โƒ— = a๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚ = a๐‘– ฬ‚ + a๐‘— ฬ‚ + a๐‘˜ ฬ‚ Also, given that magnitude of ๐‘Ÿ โƒ— is 5โˆš3 units |๐‘Ÿ โƒ— | = 5โˆš3 โˆš(๐‘Ž^2+๐‘Ž^2+๐‘Ž^2 ) = 5โˆš3 โˆš(ใ€–3๐‘Žใ€—^2 ) = 5โˆš3 โˆš3 ๐‘Ž = 5โˆš3 ๐‘Ž = 5 So, our vector ๐‘Ÿ โƒ— becomes ๐‘Ÿ โƒ— = a๐‘– ฬ‚ + a๐‘— ฬ‚ + a๐‘˜ ฬ‚ = 5๐‘– ฬ‚ + 5๐‘— ฬ‚ + 5๐‘˜ 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 9 years. He provides courses for Maths and Science at Teachoo.