Last updated at May 29, 2018 by Teachoo

Transcript

Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane ๐๏ทฏ.( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ) + 9 = 0 . The vector equation of a line passing through a point with position vector ๐๏ทฏ and parallel to vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Given, the line passes (1, 2, 3) So, ๐๏ทฏ = 1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ Finding normal of plane ๐๏ทฏ.( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ) + 9 = 0 ๐๏ทฏ. ( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ) = โ 9 โ ๐๏ทฏ. ( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ) = 9 ๐๏ทฏ. (โ1 ๐๏ทฏ โ 2 ๐๏ทฏ + 5 ๐๏ทฏ) = 9 Comparing with ๐๏ทฏ. ๐๏ทฏ = d, ๐๏ทฏ = โ ๐๏ทฏ โ 2 ๐๏ทฏ + 5 ๐๏ทฏ Since line is perpendicular to plane, the line will be parallel to the normal of the plane โด ๐๏ทฏ = ๐๏ทฏ = โ1 ๐๏ทฏ โ 2 ๐๏ทฏ + 5 ๐๏ทฏ Hence, ๐๏ทฏ = (1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐(โ1 ๐๏ทฏ โ 2 ๐๏ทฏ + 5 ๐๏ทฏ) ๐๏ทฏ = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) โ ๐ ( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ) โด Vector equation of line is ๐๏ทฏ = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) โ ๐ ( ๐๏ทฏ + 2 ๐๏ทฏ โ 5 ๐๏ทฏ)

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.