Check sibling questions

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

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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.