Chapter 10 Class 12 Vector Algebra
Ex 10.2, 9
Ex 10.2, 10 Important
Ex 10.2, 13 Important
Ex 10.2, 17 Important
Example 14 Important
Example 16 Important
Example 21 Important
Ex 10.3, 2
Ex 10.3, 3 Important
Ex 10.3, 10 Important
Ex 10.3, 13 Important
Ex 10.3, 16 Important
Example 23 Important
Example 24
Example 25 Important
Ex 10.4, 2 Important
Ex 10.4, 5 Important
Ex 10.4, 9 Important
Ex 10.4, 10
Ex 10.4, 11 (MCQ) Important
Example 28 Important
Example 29 Important
Example 30 Important You are here
Misc 6
Misc 12 Important
Misc 13
Misc 15 Important
Misc 19 (MCQ) Important
Chapter 10 Class 12 Vector Algebra
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