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

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 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