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Miscellaneous
Last updated at Feb. 4, 2020 by Teachoo
Misc 20 (Method 1) Find the vector equation of the line passing through the point (1, 2, ā4) and perpendicular to the two lines: (š„ ā 8)/3 = (š¦ + 19)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) The vector equation of a line passing through a point with position vector š ā and parallel to a vector š ā is š ā = š ā + šš ā The line passes through (1, 2, ā4) So, š ā = 1š Ģ + 2š Ģ ā 4š Ģ Given, line is perpendicular to both lines ā“ š ā is perpendicular to both lines We know that š ā Ć š ā is perpendicular to both š ā & š ā So, š ā is cross product of both lines (š„ ā 8)/3 = (š¦ + 19)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) Required normal = |ā 8(š Ģ&š Ģ&š Ģ@3&ā16&7@3&8&ā5)| = š Ģ (ā16(ā5) ā 8(7)) ā š Ģ (3(-5) ā 3(7)) + š Ģ(3(8) ā 3(ā16)) = š Ģ (80 ā 56) ā š Ģ (ā15 ā 21) + š Ģ(24 + 48) = 24š Ģ + 36š Ģ + 72š Ģ Thus, š ā = 24š Ģ + 36š Ģ + 72š Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā ā“ š ā = (1š Ģ + 2š Ģ ā 4š Ģ) + š (24š Ģ + 36š Ģ + 72š Ģ) = (š Ģ + 2š Ģ ā 4š Ģ) + š12 (2š Ģ + 3š Ģ + 6š Ģ) = (š Ģ + 2š Ģ ā 4š Ģ) + š (2š Ģ + 3š Ģ + 6š Ģ) Therefore, the equation of the line is (š Ģ + 2š Ģ ā 4š Ģ) + š (2š Ģ + 3š Ģ + 6š Ģ). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, ā4) and perpendicular to the two lines: (š„ ā 8)/3 = (š¦ + 19)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) The vector equation of a line passing through a point with position vector š ā and parallel to a vector š ā is š ā = š ā + šš ā The line passes through (1, 2, ā4) So, š ā = 1š Ģ + 2š Ģ ā 4š Ģ Let š ā = xš Ģ + yš Ģ + zš Ģ Two lines with direction ratios š1 , š1 , š1 & š2 , š2 , š2 are perpendicular if šš šš + šššš + šš šš = 0 Given, line š ā is perpendicular to (š„ ā 8)/3 = (š¦ + 19)/16 = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/( ā 5) So, 3x ā 16y + 7z = 0 and 3x + 8y ā 5z = 0 š„/(80 ā 56 ) = š¦/(21 ā ( ā15) ) = š§/(24 ā ( ā48) ) š„/(24 ) = š¦/36 = š§/72 š„/2 = š¦/3 = š§/6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, š ā = xš Ģ + yš Ģ + zš Ģ = 2kš Ģ + 3kš Ģ + 6kš Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā ā“ š ā = (š Ģ + 2š Ģ ā 4š Ģ) + š (2kš Ģ + 3kš Ģ + 6kš Ģ) = (š Ģ + 2š Ģ ā 4š Ģ) + šk (2š Ģ + 3š Ģ + 6š Ģ) = (š Ģ + 2š Ģ ā 4š Ģ) + š (2š Ģ + 3š Ģ + 6š Ģ) Therefore, the equation of the line is (š Ģ + 2š Ģ ā 4š Ģ) + š(2š Ģ + 3š Ģ + 6š Ģ)