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This question is the same as Example 27 (Supplementary NCERT)

Ex 10.5, 2 (Supplementary NCERT) - Show that a = i - 2j + 3k, b = -2i

Ex 10.5, 2 (Supplementary NCERT) - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.5, 2 (Supplementary NCERT) Show that the vectors π‘Ž βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚, 𝑏 βƒ— = βˆ’2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ and 𝑐 βƒ— = 𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚ are coplanarThree vectors π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ— are coplanar if [𝒂 βƒ—" " 𝒃 βƒ—" " 𝒄 βƒ— ] = 0 Given, 𝒂 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚ 𝒃 βƒ— = βˆ’2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ 𝒄 βƒ— = 𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + Ξ»π‘˜ Μ‚ Now, [π‘Ž βƒ—" " 𝑏 βƒ—" " 𝑐 βƒ— ] = |β– 8(1&βˆ’2&[email protected]βˆ’2&3&βˆ’[email protected]&βˆ’3&5)| = 1[(3Γ—5)βˆ’(βˆ’3Γ—βˆ’4) ] βˆ’ (βˆ’2) [(βˆ’2Γ—5)βˆ’(1Γ—βˆ’4) ] + 3[(βˆ’2Γ—βˆ’3)βˆ’(1Γ—3) ] = 1 [15βˆ’(3 Γ—4)]+2[βˆ’10βˆ’(βˆ’4)]+3[(2 Γ—3)βˆ’3] = 1 [15βˆ’12]+2[βˆ’10+4]+3[6βˆ’3] = 1 [3]+2[βˆ’6]+3[3] = 3 – 12 + 9 = 0 Since [π‘Ž βƒ—" " 𝑏 βƒ—" " 𝑐 βƒ— ] = 0 Vectors π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ— are coplanar

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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.