Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

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&7@3&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 ̂)