Last updated at May 29, 2018 by Teachoo

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 )

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

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.