Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

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 )