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Last updated at Feb. 4, 2020 by Teachoo
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Misc 20 (Method 1) Find the vector equation of the line passing through the point (1, 2, โ4) and perpendicular to the two lines: (๐ฅ โ 8)/3 = (๐ฆ + 19)/(โ16) = (๐ง โ 10)/7 and (๐ฅ โ 15)/3 = (๐ฆ โ 29)/8 = (๐ง โ 5)/(โ5) The vector equation of a line passing through a point with position vector ๐ โ and parallel to a vector ๐ โ is ๐ โ = ๐ โ + ๐๐ โ The line passes through (1, 2, โ4) So, ๐ โ = 1๐ ฬ + 2๐ ฬ โ 4๐ ฬ Given, line is perpendicular to both lines โด ๐ โ is perpendicular to both lines We know that ๐ โ ร ๐ โ is perpendicular to both ๐ โ & ๐ โ So, ๐ โ is cross product of both lines (๐ฅ โ 8)/3 = (๐ฆ + 19)/(โ16) = (๐ง โ 10)/7 and (๐ฅ โ 15)/3 = (๐ฆ โ 29)/8 = (๐ง โ 5)/(โ5) Required normal = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@3&โ16&7@3&8&โ5)| = ๐ ฬ (โ16(โ5) โ 8(7)) โ ๐ ฬ (3(-5) โ 3(7)) + ๐ ฬ(3(8) โ 3(โ16)) = ๐ ฬ (80 โ 56) โ ๐ ฬ (โ15 โ 21) + ๐ ฬ(24 + 48) = 24๐ ฬ + 36๐ ฬ + 72๐ ฬ Thus, ๐ โ = 24๐ ฬ + 36๐ ฬ + 72๐ ฬ Now, Putting value of ๐ โ & ๐ โ in formula ๐ โ = ๐ โ + ๐๐ โ โด ๐ โ = (1๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐ (24๐ ฬ + 36๐ ฬ + 72๐ ฬ) = (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐12 (2๐ ฬ + 3๐ ฬ + 6๐ ฬ) = (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐ (2๐ ฬ + 3๐ ฬ + 6๐ ฬ) Therefore, the equation of the line is (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐ (2๐ ฬ + 3๐ ฬ + 6๐ ฬ). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, โ4) and perpendicular to the two lines: (๐ฅ โ 8)/3 = (๐ฆ + 19)/(โ16) = (๐ง โ 10)/7 and (๐ฅ โ 15)/3 = (๐ฆ โ 29)/8 = (๐ง โ 5)/(โ5) The vector equation of a line passing through a point with position vector ๐ โ and parallel to a vector ๐ โ is ๐ โ = ๐ โ + ๐๐ โ The line passes through (1, 2, โ4) So, ๐ โ = 1๐ ฬ + 2๐ ฬ โ 4๐ ฬ Let ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ Two lines with direction ratios ๐1 , ๐1 , ๐1 & ๐2 , ๐2 , ๐2 are perpendicular if ๐๐ ๐๐ + ๐๐๐๐ + ๐๐ ๐๐ = 0 Given, line ๐ โ is perpendicular to (๐ฅ โ 8)/3 = (๐ฆ + 19)/16 = (๐ง โ 10)/7 and (๐ฅ โ 15)/3 = (๐ฆ โ 29)/8 = (๐ง โ 5)/( โ 5) So, 3x โ 16y + 7z = 0 and 3x + 8y โ 5z = 0 ๐ฅ/(80 โ 56 ) = ๐ฆ/(21 โ ( โ15) ) = ๐ง/(24 โ ( โ48) ) ๐ฅ/(24 ) = ๐ฆ/36 = ๐ง/72 ๐ฅ/2 = ๐ฆ/3 = ๐ง/6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ = 2k๐ ฬ + 3k๐ ฬ + 6k๐ ฬ Now, Putting value of ๐ โ & ๐ โ in formula ๐ โ = ๐ โ + ๐๐ โ โด ๐ โ = (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐ (2k๐ ฬ + 3k๐ ฬ + 6k๐ ฬ) = (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐k (2๐ ฬ + 3๐ ฬ + 6๐ ฬ) = (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐ (2๐ ฬ + 3๐ ฬ + 6๐ ฬ) Therefore, the equation of the line is (๐ ฬ + 2๐ ฬ โ 4๐ ฬ) + ๐(2๐ ฬ + 3๐ ฬ + 6๐ ฬ)
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