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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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= 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.