Check sibling questions

Misc 10 - Find unit vector parallel to parallelogram diagonal

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

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Misc 10 The two adjacent sides of a parallelogram are 2𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ + 5π‘˜ Μ‚ and 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ Find the unit vector parallel to its diagonal. Also, find its area. Let π‘Ž βƒ— and 𝑏 βƒ— are adjacent side of a parallelogram, where 𝒂 βƒ— = 2𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ + 5π‘˜ Μ‚ 𝒃 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ Let diagonal be 𝒄 βƒ— Hence, 𝒄 βƒ— = 𝒂 βƒ— + 𝒃 βƒ— = (2𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ + 5π‘˜ Μ‚) + (1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚) = (2 + 1) 𝑖 Μ‚ βˆ’ (4 + 2) 𝑗 Μ‚ + (5 βˆ’ 3)π‘˜ Μ‚ = 3π’Š Μ‚ βˆ’ 6𝒋 Μ‚ + 2π’Œ Μ‚ Magnitude of βŒˆπ‘ βƒ— βŒ‰ = √((3)^2+(βˆ’6)^2+(2)^2 ) = √(9+36+4) = √49 = 7 Unit vector in direction of 𝑐 βƒ— = 1/(π‘€π‘Žπ‘”π‘›π‘–π‘‘π‘’π‘‘π‘’ π‘œπ‘“ 𝑐 βƒ— ) Γ— 𝑐 βƒ— 𝒄 Μ‚ = 𝟏/πŸ• ("3" π’Š Μ‚" βˆ’ 6" 𝒋 Μ‚" + 2" π’Œ Μ‚ ) Finding Area of parallelogram Area of parallelogram = |π‘Ž βƒ— Γ— 𝑏 βƒ— | Now, π‘Ž βƒ— Γ— 𝑏 βƒ— = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@2&βˆ’4&5@1&βˆ’2&βˆ’3)| = 𝑖 Μ‚ (12 βˆ’ (βˆ’10) βˆ’ 𝑗 Μ‚ (βˆ’6 βˆ’5) + π‘˜ Μ‚ (βˆ’4 βˆ’ (βˆ’4)) = 22𝑖 Μ‚ + 11𝑗 Μ‚ + 0π‘˜ Μ‚ = 22𝑖 Μ‚ + 11𝑗 Μ‚ So 𝒂 βƒ— Γ— 𝒃 βƒ— = 22π’Š Μ‚ + 11𝒋 Μ‚ Now, |π‘Ž βƒ— Γ— 𝑏 βƒ— | = √(22^2+11^2 ) = √(2^2 (γ€–11)γ€—^2+11^2 ) = √( 11^2 (2^2+1)) = 11√5 Hence, Area of parallelogram = |π‘Ž βƒ— Γ— 𝑏 βƒ— |= πŸπŸβˆšπŸ“

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

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.