Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Misc 5 (Method 1) Find the vector equation of the line passing through the point (1, 2, ā4) and perpendicular to the two lines: (š„ ā 8)/3 = (š¦ + 19)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) 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 + šš ā The line passes through (1, 2, ā4) 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)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) Required normal = |ā 8(š Ģ&š Ģ&š Ģ@3&ā16&7@3&8&ā5)| = š Ģ (ā16(ā5) ā 8(7)) ā š Ģ (3(-5) ā 3(7)) + š Ģ(3(8) ā 3(ā16)) = š Ģ (80 ā 56) ā š Ģ (ā15 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 + šš 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) + š12 (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 5 (Method 2) Find the vector equation of the line passing through the point (1, 2, ā4) and perpendicular to the two lines: (š„ ā 8)/3 = (š¦ + 19)/(ā16) = (š§ ā 10)/7 and (š„ ā 15)/3 = (š¦ ā 29)/8 = (š§ ā 5)/(ā5) 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 + šš ā The line passes through (1, 2, ā4) So, š ā = 1š Ģ + 2š Ģ ā 4š Ģ Let š ā = xš Ģ + yš Ģ + zš Ģ Two lines with direction ratios š1 , š1 , š1 & š2 , š2 , š2 are perpendicular if šš šš + šššš + šš šš = 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 ( ā15) ) = š§/(24 ā ( ā48) ) 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 + šš 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) + šk (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)