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Example 30 - With reference to right handed system of mutually

Example 30 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 30 - Chapter 10 Class 12 Vector Algebra - Part 3
Example 30 - Chapter 10 Class 12 Vector Algebra - Part 4

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Example 30 If with reference to the right handed system of mutually perpendicular unit vectors 𝑖 Μ‚, 𝑗 Μ‚ and π‘˜ Μ‚, "Ξ±" βƒ— = 3𝑖 Μ‚ βˆ’ 𝑗 Μ‚, "Ξ²" βƒ— = 2𝑖 Μ‚ + 𝑗 Μ‚ – 3π‘˜ Μ‚, then express "Ξ²" Μ‚ in the form "Ξ²" βƒ— = "Ξ²" βƒ—1 + "Ξ²" βƒ—2, where "Ξ²" βƒ—1 is parallel to "Ξ±" βƒ— and "Ξ²" βƒ—2 is perpendicular to "Ξ±" βƒ—.Given 𝛼 βƒ— = 3𝑖 Μ‚ βˆ’ 𝑗 Μ‚ = 3𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 0π‘˜ Μ‚ "Ξ²" βƒ— = 2𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = 2𝑖 Μ‚ + 1𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ To show: "Ξ²" βƒ— = "Ξ²" βƒ—1 + "Ξ²" βƒ—2 Given, "Ξ²" βƒ—1 is parallel to 𝛼 βƒ— & "Ξ²" βƒ—2 is perpendicular to 𝛼 βƒ— Let "Ξ²" βƒ—1 = π€πœΆ βƒ— , πœ† being a scalar. "Ξ²" βƒ—1 = πœ† (3𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 0π‘˜ Μ‚) = 3πœ† 𝑖 Μ‚ βˆ’ πœ†π‘— Μ‚ + 0π‘˜ Μ‚ Now, "Ξ²" βƒ—2 = "Ξ²" βƒ— βˆ’ "Ξ²" βƒ—1 = ["2" 𝑖 Μ‚" + 1" 𝑗 Μ‚" βˆ’ 3" π‘˜ Μ‚ ] βˆ’ ["3" πœ†π‘– Μ‚" βˆ’ πœ†" 𝑗 Μ‚" + 0" π‘˜ Μ‚ ] = 2𝑖 Μ‚ + 1𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ βˆ’ 3πœ†π‘– Μ‚ + πœ†π‘— Μ‚ + 0π‘˜ Μ‚ = (2 βˆ’ 3πœ†) 𝑖 Μ‚ + (1 + πœ†) 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ Also, since "Ξ²" βƒ—2 is perpendicular to 𝛼 βƒ— "Ξ²" βƒ—2 . 𝜢 βƒ— = 0 ["(2 βˆ’ 3πœ†) " 𝑖 Μ‚" + (1 + πœ†) " 𝑗 Μ‚" βˆ’ 3" π‘˜ Μ‚ ]. (3𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 0π‘˜ Μ‚) = 0 (2 βˆ’ "3πœ†") Γ— 3 + (1 + πœ†) Γ— βˆ’1 + (βˆ’3) Γ— 0 = 0 6 βˆ’ 9"πœ†" βˆ’ 1 βˆ’ πœ† = 0 5 βˆ’ 10πœ† = 0 πœ† = 5/10 πœ† = 𝟏/𝟐 Putting value of πœ† in "Ξ²" βƒ—1 and "Ξ²" βƒ—2 , "Ξ²" βƒ—1 = 3πœ†π‘– Μ‚ βˆ’ πœ†π‘— Μ‚ + 0π‘˜ Μ‚ = 3. 1/2 𝑖 Μ‚ βˆ’ 1/2 𝑗 Μ‚ + 0 π‘˜ Μ‚ = πŸ‘/𝟐 π’Š Μ‚ βˆ’ 𝟏/𝟐 𝒋 Μ‚ "Ξ²" βƒ—2 = (2 βˆ’ 3πœ†) 𝑖 Μ‚ + (1 + πœ†) 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = ("2 βˆ’ 3. " 1/2) 𝑖 Μ‚ + (1+1/2) 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = 𝟏/𝟐 π’Š Μ‚ + πŸ‘/𝟐 𝒋 Μ‚βˆ’ 3π’Œ Μ‚ "Ξ²" βƒ—2 = (2 βˆ’ 3πœ†) 𝑖 Μ‚ + (1 + πœ†) 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = ("2 βˆ’ 3. " 1/2) 𝑖 Μ‚ + (1+1/2) 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = 𝟏/𝟐 π’Š Μ‚ + πŸ‘/𝟐 𝒋 Μ‚βˆ’ 3π’Œ Μ‚ Thus, "Ξ²" βƒ—1 + "Ξ²" βƒ—2 = (πŸ‘/𝟐 " " π’Š Μ‚" βˆ’ " 𝟏/𝟐 " " 𝒋 Μ‚ )+(𝟏/𝟐 " " π’Š Μ‚" + " πŸ‘/𝟐 " " 𝒋 Μ‚βˆ’" 3" π’Œ Μ‚ ) = 2𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ = "Ξ²" βƒ— Hence proved

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