Last updated at March 11, 2017 by Teachoo

Transcript

Misc 12 Let 𝑎 = 𝑖 + 4 𝑗 + 2 𝑘, 𝑏 = 3 𝑖 − 2 𝑗 + 7 𝑘 and 𝑐 = 2 𝑖 − 𝑗 + 4 𝑘 . Find a vector 𝑑 which is perpendicular to both 𝑎 and 𝑏 and 𝑐 ⋅ 𝑑 = 15 . 𝑎 = 𝑖 + 4 𝑗 + 2 𝑘 = 1 𝑖 + 4 𝑗 + 2 𝑘 𝑏 = 3 𝑖 + 2 𝑗 + 7 𝑘 𝑐 = 2 𝑖 + 𝑗 + 4 𝑘 = 2 𝑖 − 1 𝑗 + 4 𝑘 Let 𝑑 = x 𝑖 + y 𝑗 + z 𝑘 Since 𝑑 is perpendicular to 𝑎 and 𝑏 𝑑 . 𝑎 = 0 & 𝑑 . 𝑏 = 0 Also, 𝑐 . 𝑑 = 15 (2 𝑖 – 1 𝑗 + 4 𝑘). (x 𝑖 + y 𝑗 + z 𝑘) = 15 (2 × x) + (−1 × y) + (4 × 2) = 15 2x − y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 …(1) 3x − 2y + 7z = 0 …(2) & 2x − y + 4z = 15 …(3) Solving x + 4y + 2z = 0 …(1) 3x – 2y + 7z = 0 …(2) 𝑥28 − (−4) = 𝑦6 − 7 = 𝑧−2 − 12 𝑥32 = 𝑦−1 = 𝑧−14 Writing x & y in terms of z ∴ x = 32𝑧−14 = −16𝑧7 & y = −1𝑧−14 = 𝑧14 Putting values of x and y in (3) 2x – y + 4z = 15 2 −16𝑧7 − 𝑧14 + 4z = 5 −327 z − 114 z + 41 z = 15 (−64 − 1 + 56)14 z = 15 −914 z = 15 z = 15 × 14−9 z = −𝟕𝟎𝟑 Putting value of z in x & y, x = −16𝑧7 = −167 × −703 = 𝟏𝟔𝟎𝟑 y = 𝑧14 = 114 × −703 = −𝟓𝟑 Therefore, the required vector 𝑑 = x 𝑖 + y 𝑗 + z 𝑘 = 1603 𝑖 − 53 𝑗 – 703 𝑘 = 𝟏𝟑 (160 𝒊 − 5 𝒋 – 70 𝒌)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.