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Misc 14 - If a,b,c are mutually perpendicular vectors of equal

Misc 14 - Chapter 10 Class 12 Vector Algebra - Part 2 Misc 14 - Chapter 10 Class 12 Vector Algebra - Part 3


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Misc 14 If π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ— are mutually perpendicular vectors of equal magnitudes, show that the vector π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ— is equally inclined to π‘Ž βƒ—, 𝑏 βƒ— and 𝑐 βƒ— . Given π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ— are of equal magnitudes, So, |𝒂 βƒ— | = |𝒃 βƒ— | = |𝒄 βƒ— | Also, π‘Ž βƒ— , 𝑏 βƒ— , 𝑐 βƒ— are mutually perpendicular to each other So, 𝒂 βƒ— . 𝒃 βƒ— = 𝒃 βƒ— . 𝒄 βƒ— = 𝒄 βƒ— . 𝒂 βƒ— = 0 We need to show (π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ—) is equally inclined to π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ— ; (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—). 𝒂 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||π‘Ž βƒ— | cos 𝜢 where 𝛼 = angle b/w (π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ—) and π‘Ž βƒ— 𝒂 βƒ— . 𝒂 βƒ— + 𝒃 βƒ— . 𝒂 βƒ— + 𝒄 βƒ— . 𝒂 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||π‘Ž βƒ— | cos 𝛼 |π‘Ž βƒ— |2 + 0 + 0 = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||π‘Ž βƒ— | cos 𝛼 cos 𝜢 = |𝒂 βƒ— |/|𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ— | (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—). 𝒃 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑏 βƒ— | cos 𝜷 where 𝛽 = angle b/w (π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ—) and 𝑏 βƒ— 𝒂 βƒ— . 𝒃 βƒ— + 𝒃 βƒ— . 𝒃 βƒ— + 𝒄 βƒ— . 𝒃 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑏 βƒ— | cos 𝛽 0 +|𝑏 βƒ— |2 + 0 = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑏 βƒ— | cos 𝛽 cos 𝜷 = |𝒃 βƒ— |/|(𝒂 ) βƒ— + 𝒃 βƒ— + 𝒄 βƒ— | (𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ—). 𝒄 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑐 βƒ— | cos 𝜸 where 𝛾 = angle b/w = (π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ—) and 𝑐 βƒ— 𝒂 βƒ— . 𝒄 βƒ— + 𝒃 βƒ— . 𝒄 βƒ— + 𝒄 βƒ— . 𝒄 βƒ— = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑐 βƒ— | cos 𝛾 0 + 0+|𝑐 βƒ— |2 = |π‘Ž βƒ—+𝑏 βƒ—+𝑐 βƒ— ||𝑐 βƒ— | cos 𝛾 cos 𝜸 = |𝒄 βƒ— |/|𝒂 βƒ— + 𝒃 βƒ— + 𝒄 βƒ— | Property : π‘Ž βƒ— . 𝑏 βƒ— = 𝑏 βƒ— . π‘Ž βƒ— π‘Ž βƒ— . π‘Ž βƒ— = |π‘Ž βƒ— |2 Property : π‘Ž βƒ— . 𝑏 βƒ— = 𝑏 βƒ— . π‘Ž βƒ— π‘Ž βƒ— . π‘Ž βƒ— = |π‘Ž βƒ— |2 Property : π‘Ž βƒ— . 𝑏 βƒ— = 𝑏 βƒ— . π‘Ž βƒ— π‘Ž βƒ— . π‘Ž βƒ— = |π‘Ž βƒ— |2 So, cos 𝛼 = |π‘Ž βƒ— |/|π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ— | , cos 𝛽 = |𝑏 βƒ— |/|(π‘Ž ) βƒ— + 𝑏 βƒ— + 𝑐 βƒ— | , cos 𝛾 = |𝑐 βƒ— |/|π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ— | But |𝒂 βƒ— | = |𝒃 βƒ— | = |𝒄 βƒ— | Thus, cos 𝜢 = cos 𝜷 = cos 𝜸 ∴ 𝛼 = 𝛽 = 𝛾 Therefore, (π‘Ž βƒ— + 𝑏 βƒ— + 𝑐 βƒ—) is equally inclined to π‘Ž βƒ—, 𝑏 βƒ—, 𝑐 βƒ—.

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.