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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

If vector p = i + j + k and q = i – 2j + k, find a vector of magnitude

Question 22 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 22 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 22 - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 22 - CBSE Class 12 Sample Paper for 2019 Boards - Part 5

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Question 22 If 𝑝 ⃗ = 𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂ and ğ‘ž ⃗ = 𝑖 ̂ – 2𝑗 ̂ + 𝑘 ̂, find a vector of magnitude 5√3 units perpendicular to the vector ğ‘ž ⃗ and coplanar with vectors 𝑝 ⃗ and ğ‘ž ⃗ Let 𝑟 ⃗ be the required vector 𝑟 ⃗ = a𝑖 ̂ + b𝑗 ̂ + c𝑘 ̂ Now, given that 𝑟 ⃗ & ğ‘ž ⃗ are perpendicular Therefore, 𝑟 ⃗.ğ‘ž ⃗ = 0 (a𝑖 ̂ + b𝑗 ̂ + c𝑘 ̂) . (𝑖 ̂ – 2𝑗 ̂ + 𝑘 ̂) = 0 a × 1 + b × (–2) + c × 1 = 0 a – 2b + c = 0 Also, given that 𝑟 ⃗ is coplanar with vectors 𝑝 ⃗ and ğ‘ž ⃗ Therefore, [𝒑 ⃗" " 𝒒 ⃗" " 𝒓 ⃗ ] = 0 Finding [𝒑 ⃗" " 𝒒 ⃗" " 𝒓 ⃗ ] [𝑝 ⃗" " ğ‘ž ⃗" " 𝑟 ⃗ ] = |■8(1&1&1@1&−2&1@ğ‘Ž&𝑏&𝑐)| = 1[−2𝑐−𝑏 ] − 1[ğ‘âˆ’ğ‘Ž] + 1[𝑏+2ğ‘Ž] = −2𝑐−𝑏−𝑐+ğ‘Ž+𝑏+2ğ‘Ž = −3𝑐+3ğ‘Ž Since [𝑝 ⃗" " ğ‘ž ⃗" " 𝑟 ⃗ ] = 0 −3𝑐+3ğ‘Ž=0 3ğ‘Ž=3𝑐 ğ‘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 𝑟 ⃗ becomes 𝑟 ⃗ = a𝑖 ̂ + b𝑗 ̂ + c𝑘 ̂ = a𝑖 ̂ + a𝑗 ̂ + a𝑘 ̂ Also, given that magnitude of 𝑟 ⃗ is 5√3 units |𝑟 ⃗ | = 5√3 √(ğ‘Ž^2+ğ‘Ž^2+ğ‘Ž^2 ) = 5√3 √(〖3ğ‘Žã€—^2 ) = 5√3 √3 ğ‘Ž = 5√3 ğ‘Ž = 5 So, our vector 𝑟 ⃗ becomes 𝑟 ⃗ = a𝑖 ̂ + a𝑗 ̂ + a𝑘 ̂ = 5𝑖 ̂ + 5𝑗 ̂ + 5𝑘 ̂

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Davneet Singh

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.