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Ex 10.4, 2 - Find a unit vector perpendicular to a + b, a - b

Ex 10.4, 2 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.4, 2 - Chapter 10 Class 12 Vector Algebra - Part 3
Ex 10.4, 2 - Chapter 10 Class 12 Vector Algebra - Part 4

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Ex 10.4, 2 Find a unit vector perpendicular to each of the vector π‘Ž βƒ— + 𝑏 βƒ— and π‘Ž βƒ— βˆ’ 𝑏 βƒ—, where π‘Ž βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ .π‘Ž βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ 𝑏 βƒ— = 1𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ (π‘Ž βƒ— + 𝑏 βƒ—) = (3 + 1) 𝑖 Μ‚ + (2 + 2) 𝑗 Μ‚ + (2 βˆ’ 2) π‘˜ Μ‚ = 4𝑖 Μ‚ + 4𝑗 Μ‚ + 0π‘˜ Μ‚ (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) = (3 βˆ’ 1) 𝑖 Μ‚ + (2 βˆ’ 2) 𝑗 Μ‚ + (2 βˆ’ (βˆ’2)) π‘˜ Μ‚ = 2𝑖 Μ‚ + 0𝑗 Μ‚ + 4π‘˜ Μ‚ Now, we need to find a vector perpendicular to both π‘Ž βƒ— + 𝑏 βƒ— and π‘Ž βƒ— βˆ’ 𝑏 βƒ—, We know that (π‘Ž βƒ— Γ— 𝑏 βƒ—) is perpendicular to π‘Ž βƒ— and 𝑏 βƒ— Replacing π‘Ž βƒ— by (π‘Ž βƒ— + 𝑏 βƒ—) & 𝑏 βƒ— by (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) (𝒂 βƒ— + 𝒃 βƒ—) Γ— (𝒂 βƒ— βˆ’ 𝒃 βƒ—) will be perpendicular to (𝒂 βƒ— + 𝒃 βƒ—) and (𝒂 βƒ— βˆ’ 𝒃 βƒ—) Let 𝑐 βƒ— = (π‘Ž βƒ— + 𝑏 βƒ—) Γ— (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) ∴ 𝑐 βƒ— = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@β–ˆ(4@2)&β–ˆ(4@0)&β–ˆ(0@4))| = 𝑖 Μ‚ [(4Γ—4)βˆ’(0Γ—0)] βˆ’ 𝑗 Μ‚ [(4Γ—4)βˆ’(2Γ—0)] + π‘˜ Μ‚ [(4Γ—0)βˆ’(2Γ—4)] = 𝑖 Μ‚ (16 βˆ’ 0) βˆ’ 𝑗 Μ‚ (16 βˆ’ 0) + π‘˜ Μ‚ (0 βˆ’ 8) = 16 𝑖 Μ‚ βˆ’ 16𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ ∴ 𝑐 βƒ— = 16 𝑖 Μ‚ βˆ’ 16𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ Now, Unit vector of 𝑐 βƒ— = 1/(π‘šπ‘Žπ‘”π‘›π‘–π‘‘π‘’π‘‘π‘’ π‘œπ‘“π‘ βƒ— ) Γ— 𝑐 βƒ— Magnitude of 𝑐 βƒ— = √(162+(βˆ’16)2+(βˆ’8)2) |𝑐 βƒ— | = √(256+256+64) = √576 = 24 Unit vector of 𝑐 βƒ— = 1/|𝑐 βƒ— | Γ— 𝑐 βƒ— = 1/24 Γ— ["16" 𝑖 Μ‚" βˆ’ 16" 𝑗 Μ‚" βˆ’ 8" π‘˜ Μ‚ ] = 𝟐/πŸ‘ π’Š Μ‚ βˆ’ 𝟐/πŸ‘ 𝒋 Μ‚ βˆ’ 𝟏/πŸ‘ π’Œ Μ‚ . Therefore the required unit vector is 2/3 𝑖 Μ‚ βˆ’ 2/3 𝑗 Μ‚ βˆ’ 1/3 π‘˜ Μ‚ . Note: There are always two perpendicular vectors So, another vector would be = βˆ’(𝟐/πŸ‘ " " π’Š Μ‚" βˆ’ " 𝟐/πŸ‘ " " 𝒋 Μ‚" βˆ’ " 𝟏/πŸ‘ " " π’Œ Μ‚ ) = (βˆ’πŸ)/πŸ‘ π’Š Μ‚ + 𝟐/πŸ‘ 𝒋 Μ‚ + 𝟏/πŸ‘ π’Œ Μ‚ Hence, the perpendicular vectors are 2/3 " " 𝑖 Μ‚" βˆ’ " 2/3 " " 𝑗 Μ‚" βˆ’ " 1/3 " " π‘˜ Μ‚ & (βˆ’2)/3 𝑖 Μ‚ + 2/3 𝑗 Μ‚ + 1/3 π‘˜ Μ‚

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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.