Example 26 - Write all unit vectors in XY-plane - Class 12 Vector

Example 26 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 26 - Chapter 10 Class 12 Vector Algebra - Part 3
Example 26 - Chapter 10 Class 12 Vector Algebra - Part 4

  1. Chapter 10 Class 12 Vector Algebra (Term 2)
  2. Serial order wise

Transcript

= 𝑖 Μ‚(1 βˆ’ (-4)) βˆ’ j Μ‚ (3 βˆ’ 4) + π‘˜ Μ‚ (βˆ’3 βˆ’1) = 𝑖 Μ‚(1 + 4) βˆ’ j Μ‚ (βˆ’1) + π‘˜ Μ‚ (βˆ’4) = 5π’Š Μ‚ + 𝒋 Μ‚ βˆ’ 4π’Œ Μ‚ Magnitude of π‘Ž βƒ— Γ— 𝑏 βƒ— = √(52+1^2+(βˆ’4)2) |𝒂 βƒ— Γ— 𝒃 βƒ— | = √(25+1+16) = βˆšπŸ’πŸ Area of parallelogram ABCD = |π‘Ž βƒ— Γ— 𝑏 βƒ— | = √42 Therefore, the required area is βˆšπŸ’πŸ . Let the unit vector be 𝒂 βƒ— We know that π‘Ž βƒ— = π‘₯𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚ Since the vector is in XY plane, there is no Z –coordinate. Hence, π‘Ž βƒ— = x𝑖 Μ‚ + y𝑗 Μ‚ + 0π’Œ Μ‚ 𝒂 βƒ— = π’™π’Š Μ‚ + y𝒋 Μ‚ Taking a general vector π‘Ž βƒ—, Making an angle 𝛉 with the x – axis Unit vector in direction of x axis is 𝑖 Μ‚ & in y axis is 𝑗 Μ‚ Angle with X-axis Since π‘Ž βƒ— makes an angle of ΞΈ with x-axis So, angle between 𝒂 βƒ— & π’Š Μ‚ is ΞΈ We know that, π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ, Putting π‘Ž βƒ— = π‘Ž βƒ— , 𝑏 βƒ— = 𝑖 Μ‚ & ΞΈ = ΞΈ 𝒂 βƒ— .π’Š Μ‚ = |𝒂 βƒ— ||π’Š Μ‚ | cos ΞΈ π‘Ž βƒ— .𝑖 Μ‚ = 1 Γ— 1 Γ— cos ΞΈ π‘Ž βƒ— . 𝑖 Μ‚ = cos ΞΈ (π‘₯𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚). 𝑖 Μ‚ = cos ΞΈ (π‘₯𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚). (1𝑖 Μ‚ + 0𝑗 Μ‚ + 0π‘˜ Μ‚) = cos ΞΈ π‘₯.1 + y.0 + 0.0 = cos ΞΈ (As π‘Ž βƒ— is unit vector, |π‘Ž βƒ— | = 1 & 𝑖 Μ‚ is a unit vector, |𝑖 Μ‚ | = 1) x = cos ΞΈ Angle with Y-axis π‘Ž βƒ— makes an angle of (90Β° – ΞΈ) with y-axis So, angle between 𝒂 βƒ— & 𝒋 Μ‚ is (90Β° – ΞΈ) We know that, π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ, Putting π‘Ž βƒ— = π‘Ž βƒ— , 𝑏 βƒ— = 𝑗 Μ‚ & ΞΈ = (90Β° – ΞΈ) 𝒂 βƒ— .𝒋 Μ‚ = |𝒂 βƒ— ||𝒋 Μ‚ | cos (90Β° – ΞΈ) π‘Ž βƒ— .𝑗 Μ‚ = 1 Γ— 1 Γ— cos (90Β° – ΞΈ) π‘Ž βƒ— .𝑗 Μ‚ = cos (90Β° – ΞΈ) 𝒂 βƒ— .𝒋 Μ‚ = sin ΞΈ (As π‘Ž βƒ— is unit vector, |π‘Ž βƒ— | = 1 & 𝑗 Μ‚ is a unit vector, |𝑗 Μ‚ | = 1) (π‘₯𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚). 𝑗 Μ‚ = sin ΞΈ (π‘₯𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚). (0𝑖 Μ‚ + 1𝑗 Μ‚ + 0π‘˜ Μ‚) = sin ΞΈ π‘₯.0 + y.1 + 0.0 = sin ΞΈ y = sin ΞΈ Thus, π‘Ž βƒ— = x𝑖 Μ‚ + y𝑗 Μ‚ 𝒂 βƒ— = cos πœƒπ’Š Μ‚ + sin πœƒ 𝒋 Μ‚ This value will be true in all quadrants So, 0 ≀ ΞΈ ≀ 2Ο€ Therefore, 𝒂 βƒ— = cos πœƒπ’Š Μ‚ + sinπœƒπ’‹ Μ‚ ; for 0 ≀ ΞΈ ≀ 2Ο€

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.