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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Question 22 If 𝑝 βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚ and π‘ž βƒ— = 𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚, find a vector of magnitude 5√3 units perpendicular to the vector π‘ž βƒ— and coplanar with vectors 𝑝 βƒ— and π‘ž βƒ— Let π‘Ÿ βƒ— be the required vector π‘Ÿ βƒ— = a𝑖 Μ‚ + b𝑗 Μ‚ + cπ‘˜ Μ‚ Now, given that π‘Ÿ βƒ— & π‘ž βƒ— are perpendicular Therefore, π‘Ÿ βƒ—.π‘ž βƒ— = 0 (a𝑖 Μ‚ + b𝑗 Μ‚ + cπ‘˜ Μ‚) . (𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚) = 0 a Γ— 1 + b Γ— (–2) + c Γ— 1 = 0 a – 2b + c = 0 Also, given that π‘Ÿ βƒ— is coplanar with vectors 𝑝 βƒ— and π‘ž βƒ— Therefore, [𝒑 βƒ—" " 𝒒 βƒ—" " 𝒓 βƒ— ] = 0 Finding [𝒑 βƒ—" " 𝒒 βƒ—" " 𝒓 βƒ— ] [𝑝 βƒ—" " π‘ž βƒ—" " π‘Ÿ βƒ— ] = |β– 8(1&1&1@1&βˆ’2&1@π‘Ž&𝑏&𝑐)| = 1[βˆ’2π‘βˆ’π‘ ] βˆ’ 1[π‘βˆ’π‘Ž] + 1[𝑏+2π‘Ž] = βˆ’2π‘βˆ’π‘βˆ’π‘+π‘Ž+𝑏+2π‘Ž = βˆ’3𝑐+3π‘Ž Since [𝑝 βƒ—" " π‘ž βƒ—" " π‘Ÿ βƒ— ] = 0 βˆ’3𝑐+3π‘Ž=0 3π‘Ž=3𝑐 π‘Ž=𝑐 Now, from (1) a – 2b + c = 0 Putting c = a a – 2b + a = 0 2a – 2b = 0 2a = 2b a = b Therefore, a = b = c So, our vector π‘Ÿ βƒ— becomes π‘Ÿ βƒ— = a𝑖 Μ‚ + b𝑗 Μ‚ + cπ‘˜ Μ‚ = a𝑖 Μ‚ + a𝑗 Μ‚ + aπ‘˜ Μ‚ Also, given that magnitude of π‘Ÿ βƒ— is 5√3 units |π‘Ÿ βƒ— | = 5√3 √(π‘Ž^2+π‘Ž^2+π‘Ž^2 ) = 5√3 √(γ€–3π‘Žγ€—^2 ) = 5√3 √3 π‘Ž = 5√3 π‘Ž = 5 So, our vector π‘Ÿ βƒ— becomes π‘Ÿ βƒ— = a𝑖 Μ‚ + a𝑗 Μ‚ + aπ‘˜ Μ‚ = 5𝑖 Μ‚ + 5𝑗 Μ‚ + 5π‘˜ Μ‚

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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.