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Ex 10.3, 12 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.3, 12 - Chapter 10 Class 12 Vector Algebra - Part 3

Ex 10.3, 12 - Chapter 10 Class 12 Vector Algebra - Part 4

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Ex 10.3, 12 (Introduction) If π‘Ž βƒ— .π‘Ž βƒ— = 0 & π‘Ž βƒ— . 𝑏 βƒ— = 0, then what can be concluded about the vector 𝑏 βƒ—? π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— | |𝑏 βƒ— | cos ΞΈ , ΞΈ in the angle b/w π‘Ž βƒ— and 𝑏 βƒ— Let π‘Ž βƒ— = 0 βƒ— = 0𝑖 Μ‚ + 0𝑗 Μ‚ + 0π‘˜ Μ‚ 𝒂 βƒ— = 0π’Š Μ‚ + 0𝒋 Μ‚ + 0π’Œ Μ‚ Let 𝒃 βƒ— = 8π’Š Μ‚ – 5𝒋 Μ‚ + 2π’Œ Μ‚ π‘Ž βƒ—. 𝑏 βƒ— = 0.8 + 0.(–5) + 0.2 = 0 + 0 + 0 = 0 So, π‘Ž βƒ—. 𝑏 βƒ— = 0 𝒂 βƒ— = 0π’Š Μ‚ + 0𝒋 Μ‚ + 0π’Œ Μ‚ Let 𝒃 βƒ— = 3π’Š Μ‚ – 4𝒋 Μ‚ + 7π’Œ Μ‚ π‘Ž βƒ—. 𝑏 βƒ— = 0.3 + 0(βˆ’4) + 0.7 = 0 + 0 + 0 = 0 So, π‘Ž βƒ—. 𝑏 βƒ— = 0 Hence, if π‘Ž βƒ— = 0 βƒ—, then π‘Ž βƒ—.𝑏 βƒ— = 0 for any vector 𝑏 βƒ— Ex 10.3, 12 If π‘Ž βƒ— .π‘Ž βƒ— = 0 and π‘Ž βƒ— . 𝑏 βƒ— = 0, then what can be concluded about the vector 𝑏 βƒ— ? Given, π‘Ž βƒ—. π‘Ž βƒ— = 0 |π‘Ž βƒ— | |π‘Ž βƒ— | cos 0 = 0 |π‘Ž βƒ— |2 cos 0 = 0 |π‘Ž βƒ— |2 Γ— 1 = 0 |π‘Ž βƒ— |2 = 0 |π‘Ž βƒ— | = 0 So, π‘Ž βƒ— = 0 βƒ— Now it is given, π‘Ž βƒ— . 𝑏 βƒ— = 0 0 βƒ— . 𝑏 βƒ— = 0 is true, for any vector 𝑏 βƒ— Therefore, if π‘Ž βƒ— . π‘Ž βƒ— = 0 and π‘Ž βƒ— . 𝑏 βƒ— = 0, then π‘Ž βƒ— = 0 βƒ— and 𝒃 βƒ— can be any vector

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.