Misc 20 - Find vector equation of line perpendicular to two - Miscellaneous

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  1. Class 12
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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﷯ + 𝜆 𝒃﷯ The line passes through (1,2, −4) So, 𝑎﷯ = 1 𝑖﷯ + 2 𝑗﷯ − 4 𝑘﷯ 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 = 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) = 24 𝑖﷯ + 36 𝑗﷯ + 72 𝑘﷯ Thus, 𝑏﷯ = 24 𝑖﷯ + 36 𝑗﷯ + 72 𝑘﷯ Now, Putting value of 𝑎﷯ & 𝑏﷯ in formula 𝑟﷯ = 𝑎﷯ + 𝜆 𝑏﷯ ∴ 𝑟﷯ = (1 𝑖﷯ + 2 𝑗﷯ – 4 𝑘﷯) + 𝜆 (24 𝑖﷯ + 36 𝑗﷯ + 72 𝑘﷯) = ( 𝑖﷯ + 2 𝑗﷯ – 4 𝑘﷯) + 𝜆12 (2 𝑖﷯ + 3 𝑗﷯ + 6 𝑘﷯) = ( 𝑖﷯ + 2 𝑗﷯ – 4 𝑘﷯) + 𝜆 (2 𝑖﷯ + 3 𝑗﷯ + 6 𝑘﷯) Therefore, the equation of the line is ( 𝒊﷯ + 2 𝒋﷯ – 4 𝒌﷯) + 𝜆 (2 𝒊﷯ + 3 𝒋﷯ + 6 𝒌﷯). 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﷯ + 𝜆 𝒃﷯ The line passes through (1,2, −4) So, 𝑎﷯ = 1 𝑖﷯ + 2 𝑗﷯ − 4 𝑘﷯ Let 𝑏﷯ = x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ Two lines with direction ratios 𝑎1 , 𝑏1 , 𝑐1 and 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﷯ = x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ = 2k 𝑖﷯ + 3k 𝑗﷯ + 6k 𝑘﷯ Now, Putting value of 𝑎﷯ & 𝑏﷯ in formula 𝑟﷯ = 𝑎﷯ + 𝜆 𝑏﷯ ∴ 𝑟﷯ = ( 𝑖﷯ + 2 𝑗﷯ − 4 𝑘﷯) + 𝜆 (2k 𝑖﷯ + 3k 𝑗﷯ + 6k 𝑘﷯) = ( 𝑖﷯ + 2 𝑗﷯ − 4 𝑘﷯) + 𝜆k (2 𝑖﷯ + 3 𝑗﷯ + 6 𝑘﷯) = ( 𝑖﷯ + 2 𝑗﷯ − 4 𝑘﷯) + 𝜆 (2 𝑖﷯ + 3 𝑗﷯ + 6 𝑘﷯) Therefore, the equation of the line is ( 𝒊﷯ + 2 𝒋﷯ − 4 𝒌﷯) + 𝜆(2 𝒊﷯ + 3 𝒋﷯ + 6 𝒌﷯)

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