Examples

Example 1
Important

Example 2 Important

Example 3 Important

Example 4

Example 5

Example 6 Important

Example 7

Example 8 Important

Example 9

Example 10

Example 11 (i) Important

Example 11 (ii) Important

Example 12 Important

Example 13

Example 14 Important

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important

Example 29 Important

Example 30 Important You are here

Example 26 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Last updated at April 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