







Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7 Deleted for CBSE Board 2023 Exams
Misc 8 Important Deleted for CBSE Board 2023 Exams
Misc 9 Important
Misc 10 Deleted for CBSE Board 2023 Exams
Misc 11 Important Deleted for CBSE Board 2023 Exams
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc 13 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important Deleted for CBSE Board 2023 Exams
Misc 15 Important Deleted for CBSE Board 2023 Exams
Misc 16 Deleted for CBSE Board 2023 Exams
Misc 17 Important Deleted for CBSE Board 2023 Exams
Misc 18 Important Deleted for CBSE Board 2023 Exams
Misc 19 Deleted for CBSE Board 2023 Exams
Misc 20 Important You are here
Misc 21 Important Deleted for CBSE Board 2023 Exams
Misc 22 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 23 (MCQ) Important Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at March 22, 2023 by Teachoo
Misc 20 (Method 1) Find the vector equation of the line passing through the point (1, 2, –4) and perpendicular to the two lines: (𝑥 − 8)/3 = (𝑦 + 19)/(−16) = (𝑧 − 10)/7 and (𝑥 − 15)/3 = (𝑦 − 29)/8 = (𝑧 − 5)/(−5) The vector equation of a line passing through a point with position vector 𝑎 ⃗ and parallel to a vector 𝑏 ⃗ is 𝒓 ⃗ = 𝒂 ⃗ + 𝜆𝒃 ⃗ The line passes through (1, 2, −4) So, 𝑎 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ Given, line is perpendicular to both lines ∴ 𝑏 ⃗ is perpendicular to both lines We know that 𝑎 ⃗ × 𝑏 ⃗ is perpendicular to both 𝑎 ⃗ & 𝑏 ⃗ So, 𝑏 ⃗ is cross product of both lines (𝑥 − 8)/3 = (𝑦 + 19)/(−16) = (𝑧 − 10)/7 and (𝑥 − 15)/3 = (𝑦 − 29)/8 = (𝑧 − 5)/(−5) Required normal = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@3&−16&[email protected]&8&−5)| = 𝑖 ̂ (–16(−5) – 8(7)) – 𝑗 ̂ (3(-5) – 3(7)) + 𝑘 ̂(3(8) – 3(–16)) = 𝑖 ̂ (80 – 56) – 𝑗 ̂ (–15 – 21) + 𝑘 ̂(24 + 48) = 24𝑖 ̂ + 36𝑗 ̂ + 72𝑘 ̂ Thus, 𝑏 ⃗ = 24𝑖 ̂ + 36𝑗 ̂ + 72𝑘 ̂ Now, Putting value of 𝑎 ⃗ & 𝑏 ⃗ in formula 𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗ ∴ 𝑟 ⃗ = (1𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂) + 𝜆 (24𝑖 ̂ + 36𝑗 ̂ + 72𝑘 ̂) = (𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂) + 𝜆12 (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂) = (𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂) + 𝜆 (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂) Therefore, the equation of the line is (𝒊 ̂ + 2𝒋 ̂ – 4𝒌 ̂) + 𝜆 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, –4) and perpendicular to the two lines: (𝑥 − 8)/3 = (𝑦 + 19)/(−16) = (𝑧 − 10)/7 and (𝑥 − 15)/3 = (𝑦 − 29)/8 = (𝑧 − 5)/(−5) The vector equation of a line passing through a point with position vector 𝑎 ⃗ and parallel to a vector 𝑏 ⃗ is 𝒓 ⃗ = 𝒂 ⃗ + 𝜆𝒃 ⃗ The line passes through (1, 2, −4) So, 𝑎 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ Let 𝑏 ⃗ = x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ Two lines with direction ratios 𝑎1 , 𝑏1 , 𝑐1 & 𝑎2 , 𝑏2 , 𝑐2 are perpendicular if 𝒂𝟏 𝒂𝟐 + 𝒃𝟏𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 0 Given, line 𝑏 ⃗ is perpendicular to (𝑥 − 8)/3 = (𝑦 + 19)/16 = (𝑧 − 10)/7 and (𝑥 − 15)/3 = (𝑦 − 29)/8 = (𝑧 − 5)/( − 5) So, 3x − 16y + 7z = 0 and 3x + 8y − 5z = 0 𝑥/(80 − 56 ) = 𝑦/(21 − ( −15) ) = 𝑧/(24 − ( −48) ) 𝑥/(24 ) = 𝑦/36 = 𝑧/72 𝑥/2 = 𝑦/3 = 𝑧/6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, 𝑏 ⃗ = x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ = 2k𝑖 ̂ + 3k𝑗 ̂ + 6k𝑘 ̂ Now, Putting value of 𝑎 ⃗ & 𝑏 ⃗ in formula 𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗ ∴ 𝑟 ⃗ = (𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) + 𝜆 (2k𝑖 ̂ + 3k𝑗 ̂ + 6k𝑘 ̂) = (𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) + 𝜆k (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂) = (𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) + 𝜆 (2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂) Therefore, the equation of the line is (𝒊 ̂ + 2𝒋 ̂ − 4𝒌 ̂) + 𝜆(2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂)