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Example 23 - Find a unit vector perpendicular to a + b, a - b

Example 23 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 23 - Chapter 10 Class 12 Vector Algebra - Part 3

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Example 23 - Chapter 10 Class 12 Vector Algebra - Part 4

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Example 23 Find a unit vector perpendicular to each of the vectors π‘Ž βƒ— + 𝑏 βƒ— and π‘Ž βƒ— βˆ’ 𝑏 βƒ— where π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚, b = 𝑖 Μ‚ + 2 𝑗 Μ‚ + 3π‘˜ Μ‚ . Finding (𝒂 βƒ— + 𝒃 βƒ—) and (𝒂 βƒ— βˆ’ 𝒃 βƒ—) (𝒂 βƒ— + 𝒃 βƒ—) = (1 + 1) 𝑖 Μ‚ + (1 + 2) 𝑗 Μ‚ + (1 + 3) π‘˜ Μ‚ = 2π’Š Μ‚ + 3𝒋 Μ‚ + 4π’Œ Μ‚ (𝒂 βƒ— βˆ’ 𝒃 βƒ—) = (1 βˆ’ 1) 𝑖 Μ‚ + (1 βˆ’ 2) 𝑗 Μ‚ + (1 βˆ’ 3) π‘˜ Μ‚ = 0π’Š Μ‚ βˆ’ 1𝒋 Μ‚ βˆ’ 2π’Œ 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 (π‘Ž βƒ— + 𝑏 βƒ—) and (π‘Ž 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‚@2&3&4@0&βˆ’1&βˆ’2)| = 𝑖 Μ‚ [(3Γ—βˆ’2)βˆ’(βˆ’1Γ—4)] βˆ’π‘— Μ‚ [(2Γ—βˆ’2)βˆ’(0Γ—4)] + π‘˜ Μ‚ [(2Γ—βˆ’1)βˆ’(0Γ—3)] = 𝑖 Μ‚ [βˆ’6βˆ’(βˆ’4)] βˆ’π‘— Μ‚ [βˆ’4βˆ’0] + π‘˜ Μ‚ [βˆ’2βˆ’0] = 𝑖 Μ‚ (βˆ’6 + 4) βˆ’π‘— Μ‚ (βˆ’4) + π‘˜ Μ‚(βˆ’2) = βˆ’2π’Š Μ‚ + 4𝒋 Μ‚ βˆ’ 2π’Œ Μ‚ Since we need to find unit vector perpendicular Unit vector of 𝑐 βƒ— = 𝟏/(π‘΄π’‚π’ˆπ’π’Šπ’•π’–π’…π’† 𝒐𝒇𝒄 βƒ— ) Γ— 𝒄 βƒ— = 1/√((βˆ’2)2 + (4)^2 + (βˆ’2)2) Γ— (βˆ’2𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚) = 1/√(4 + 16 + 4) Γ— (βˆ’2𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚) = 1/(2√6) Γ— (βˆ’2𝑖 Μ‚ + 4𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚) = (βˆ’πŸ)/βˆšπŸ” π’Š Μ‚ + 𝟐/βˆšπŸ” 𝒋 Μ‚ βˆ’ 𝟏/βˆšπŸ” π’Œ Μ‚ Note: There are always two perpendicular vectors So, another vector would be = βˆ’((βˆ’1)/√6 𝑖 Μ‚" + " 2/√6 𝑗 Μ‚" βˆ’ " 1/√6 π‘˜ Μ‚ ) = 𝟏/βˆšπŸ” π’Š Μ‚" βˆ’" 𝟐/βˆšπŸ” 𝒋 Μ‚" + " 𝟏/βˆšπŸ” π’Œ Μ‚ Hence, Perpendicular vectors are (βˆ’1)/√6 𝑖 Μ‚ + 2/√6 𝑗 Μ‚ βˆ’ 1/√6 π‘˜ Μ‚ & 1/√6 𝑖 Μ‚" βˆ’" 2/√6 𝑗 Μ‚" + " 1/√6 π‘˜ Μ‚

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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.