


Chapter 10 Class 12 Vector Algebra
Chapter 10 Class 12 Vector Algebra
Last updated at Dec. 16, 2024 by Teachoo
Example 30 If with reference to the right handed system of mutually perpendicular unit vectors 𝑖 ̂, 𝑗 ̂ and 𝑘 ̂, "α" ⃗ = 3𝑖 ̂ − 𝑗 ̂, "β" ⃗ = 2𝑖 ̂ + 𝑗 ̂ – 3𝑘 ̂, 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" ⃗.Given 𝛼 ⃗ = 3𝑖 ̂ − 𝑗 ̂ = 3𝑖 ̂ − 𝑗 ̂ + 0𝑘 ̂ "β" ⃗ = 2𝑖 ̂ + 𝑗 ̂ − 3𝑘 ̂ = 2𝑖 ̂ + 1𝑗 ̂ − 3𝑘 ̂ To show: "β" ⃗ = "β" ⃗1 + "β" ⃗2 Given, "β" ⃗1 is parallel to 𝛼 ⃗ & "β" ⃗2 is perpendicular to 𝛼 ⃗ Let "β" ⃗1 = 𝝀𝜶 ⃗ , 𝜆 being a scalar. "β" ⃗1 = 𝜆 (3𝑖 ̂ − 1𝑗 ̂ + 0𝑘 ̂) = 3𝜆 𝑖 ̂ − 𝜆𝑗 ̂ + 0𝑘 ̂ Now, "β" ⃗2 = "β" ⃗ − "β" ⃗1 = ["2" 𝑖 ̂" + 1" 𝑗 ̂" − 3" 𝑘 ̂ ] − ["3" 𝜆𝑖 ̂" − 𝜆" 𝑗 ̂" + 0" 𝑘 ̂ ] = 2𝑖 ̂ + 1𝑗 ̂ − 3𝑘 ̂ − 3𝜆𝑖 ̂ + 𝜆𝑗 ̂ + 0𝑘 ̂ = (2 − 3𝜆) 𝑖 ̂ + (1 + 𝜆) 𝑗 ̂ − 3𝑘 ̂ Also, since "β" ⃗2 is perpendicular to 𝛼 ⃗ "β" ⃗2 . 𝜶 ⃗ = 0 ["(2 − 3𝜆) " 𝑖 ̂" + (1 + 𝜆) " 𝑗 ̂" − 3" 𝑘 ̂ ]. (3𝑖 ̂ − 1𝑗 ̂ + 0𝑘 ̂) = 0 (2 − "3𝜆") × 3 + (1 + 𝜆) × −1 + (−3) × 0 = 0 6 − 9"𝜆" − 1 − 𝜆 = 0 5 − 10𝜆 = 0 𝜆 = 5/10 𝜆 = 𝟏/𝟐 Putting value of 𝜆 in "β" ⃗1 and "β" ⃗2 , "β" ⃗1 = 3𝜆𝑖 ̂ − 𝜆𝑗 ̂ + 0𝑘 ̂ = 3. 1/2 𝑖 ̂ − 1/2 𝑗 ̂ + 0 𝑘 ̂ = 𝟑/𝟐 𝒊 ̂ − 𝟏/𝟐 𝒋 ̂ "β" ⃗2 = (2 − 3𝜆) 𝑖 ̂ + (1 + 𝜆) 𝑗 ̂ − 3𝑘 ̂ = ("2 − 3. " 1/2) 𝑖 ̂ + (1+1/2) 𝑗 ̂ − 3𝑘 ̂ = 𝟏/𝟐 𝒊 ̂ + 𝟑/𝟐 𝒋 ̂− 3𝒌 ̂ "β" ⃗2 = (2 − 3𝜆) 𝑖 ̂ + (1 + 𝜆) 𝑗 ̂ − 3𝑘 ̂ = ("2 − 3. " 1/2) 𝑖 ̂ + (1+1/2) 𝑗 ̂ − 3𝑘 ̂ = 𝟏/𝟐 𝒊 ̂ + 𝟑/𝟐 𝒋 ̂− 3𝒌 ̂ Thus, "β" ⃗1 + "β" ⃗2 = (𝟑/𝟐 " " 𝒊 ̂" − " 𝟏/𝟐 " " 𝒋 ̂ )+(𝟏/𝟐 " " 𝒊 ̂" + " 𝟑/𝟐 " " 𝒋 ̂−" 3" 𝒌 ̂ ) = 2𝑖 ̂ + 𝑗 ̂ − 3𝑘 ̂ = "β" ⃗ Hence proved