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Example 30 - With reference to right handed system of mutually - Scalar product - Defination

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Slide81.JPG

  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 30 If with reference to the right handed system of mutually perpendicular unit vectors 𝑖﷯, 𝑗﷯ and 𝑘﷯, α﷯ = 3 𝑖﷯ − 𝑗﷯, β﷯ = 2 𝑖﷯ + 𝑗﷯ – 3 𝑘﷯, then express β﷯ in the form β﷯ = β﷯1 + β﷯2, where β﷯1 is parallel to α﷯ and β﷯2 is perpendicular to α﷯. 𝛼﷯ = 3 𝑖﷯ − 𝑗﷯ = 3 𝑖﷯ − 𝑗﷯ + 0 𝑘﷯ β﷯ = 2 𝑖﷯ + 𝑗﷯ − 3 𝑘﷯ = 2 𝑖﷯ + 1 𝑗﷯ − 3 𝑘﷯ To show: β﷯ = β﷯1 + β﷯2 Given, β﷯1 is parallel to 𝛼﷯ & β﷯2 is perpendicular to 𝛼﷯ Let β﷯1 = 𝝀 𝜶﷯ , 𝜆 being a scalar. β﷯1 = 𝜆 (3 𝑖﷯ − 1 𝑗﷯ + 0 𝑘﷯) = 3𝜆 𝑖﷯ − 𝜆 𝑗﷯ + 0 𝑘﷯ Now, β﷯2 = β﷯ − β﷯1 = 2 𝑖﷯ + 1 𝑗﷯ − 3 𝑘﷯﷯ − 3𝜆 𝑖﷯ − 𝜆 𝑗﷯ + 0 𝑘﷯﷯ = 2 𝑖﷯ + 1 𝑗﷯ − 3 𝑘﷯ − 3𝜆 𝑖﷯ + 𝜆 𝑗﷯ + 0 𝑘﷯ = (2 − 3𝜆) 𝑖﷯ + (1 + 𝜆) 𝑗﷯ − 3 𝑘﷯ Also, since β﷯2 is perpendicular to 𝛼﷯ So, β﷯2 . 𝜶﷯ = 0 (2 − 3𝜆) 𝑖﷯ + (1 + 𝜆) 𝑗﷯ − 3 𝑘﷯﷯. (3 𝑖﷯ − 1 𝑗﷯ + 0 𝑘﷯) = 0 (2 − 3𝜆) × 3 + (1 + 𝜆) × −1 + (−3) × 0 = 0 6 − 9𝜆 − 1 − 𝜆 = 0 5 − 10𝜆 = 0 10𝜆 = 5 𝜆 = 5﷮10﷯ 𝜆 = 𝟏﷮𝟐﷯ Putting value of 𝜆 in β﷯1 and β﷯2 , β﷯1 = 3𝜆 𝑖﷯ − 𝜆 𝑗﷯ + 0 𝑘﷯ = 3. 1﷮2﷯ 𝑖﷯ − 1﷮2﷯ 𝑗﷯ + 0 𝑘﷯ = 𝟑﷮𝟐﷯ 𝒊﷯ − 𝟏﷮𝟐﷯ 𝒋﷯ β﷯2 = (2 − 3𝜆) 𝑖﷯ + (1 + 𝜆) 𝑗﷯ − 3 𝑘﷯ = 2 − 3. 1﷮2﷯﷯ 𝑖﷯ + 1+ 1﷮2﷯﷯ 𝑗﷯ − 3 𝑘﷯ = 𝟏﷮𝟐﷯ 𝒊﷯ + 𝟑﷮𝟐﷯ 𝒋﷯− 3 𝒌﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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