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

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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: 8 3 = + 19 16 = 10 7 and 15 3 = 29 8 = 5 5 The vector equation of a line passing through a point with position vector and parallel to a vector is = + The line passes through (1,2, 4) So, = 1 + 2 4 Given, line is perpendicular to both lines is perpendicular to both lines We know that is perpendicular to both & So, is cross product of both lines 8 3 = + 19 16 = 10 7 and 15 3 = 29 8 = 5 5 Required normal = 3 16 7 3 8 5 = ( 16(-5) 8(7)) (3(-5) 3(7)) + (3(8) 3( 16)) = (80 56) ( 15 21) + (24 + 48) = 24 + 36 + 72 Thus, = 24 + 36 + 72 Now, Putting value of & in formula = + = (1 + 2 4 ) + (24 + 36 + 72 ) = ( + 2 4 ) + 12 (2 + 3 + 6 ) = ( + 2 4 ) + (2 + 3 + 6 ) Therefore, the equation of the line is ( + 2 4 ) + (2 + 3 + 6 ). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, 4) and perpendicular to the two lines: 8 3 = + 19 16 = 10 7 and 15 3 = 29 8 = 5 5 The vector equation of a line passing through a point with position vector and parallel to a vector is = + The line passes through (1,2, 4) So, = 1 + 2 4 Let = x + y + z Two lines with direction ratios 1 , 1 , 1 and 2 , 2 , 2 are perpendicular if + + = 0 Given, line is perpendicular to 8 3 = + 19 16 = 10 7 and 15 3 = 29 8 = 5 5 So, 3x 16y + 7z = 0 and 3x + 8y 5z = 0 80 56 = 21 ( 15) = 24 ( 48) 24 = 36 = 72 2 = 3 = 6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, = x + y + z = 2k + 3k + 6k Now, Putting value of & in formula = + = ( + 2 4 ) + (2k + 3k + 6k ) = ( + 2 4 ) + k (2 + 3 + 6 ) = ( + 2 4 ) + (2 + 3 + 6 ) Therefore, the equation of the line is ( + 2 4 ) + (2 + 3 + 6 )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.