Last updated at May 29, 2018 by Teachoo

Transcript

Misc 19 (Method 1) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes ๐๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐๏ทฏ and parallel to a vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Given, the line passes through (1, 2, 3) So, ๐๏ทฏ = 1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ Given, line is parallel to both planes โด Line is perpendicular to normal of both planes. i.e. ๐๏ทฏ is perpendicular to normal of both planes. We know that ๐๏ทฏ ร ๐๏ทฏ is perpendicular to both ๐๏ทฏ & ๐๏ทฏ So, ๐๏ทฏ is cross product of normals of planes ๐๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 Required normal = ๐๏ทฏ๏ทฎ ๐๏ทฏ๏ทฎ ๐๏ทฏ๏ทฎ1๏ทฎโ1๏ทฎ2๏ทฎ3๏ทฎ1๏ทฎ1๏ทฏ๏ทฏ = ๐๏ทฏ (โ1(1) โ 1(2)) โ ๐๏ทฏ (1(1) โ 3(2)) + ๐๏ทฏ(1(1) โ 3(โ1)) = ๐๏ทฏ (โ1 โ 2) โ ๐๏ทฏ (1 โ 6) + ๐๏ทฏ(1 + 3) = โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ Thus, ๐๏ทฏ = โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ Now, Putting value of ๐๏ทฏ & ๐๏ทฏ in formula ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) Therefore, the equation of the line is ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ). Misc 19 (Method 2) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes ๐๏ทฏ . ( ๐๏ทฏ โ ๐๏ทฏ + 2 ๐๏ทฏ) = 5 and ๐๏ทฏ . (3 ๐๏ทฏ + ๐๏ทฏ + ๐๏ทฏ) = 6 . The vector equation of a line passing through a point with position vector ๐๏ทฏ and parallel to a vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Given, the line passes through (1, 2, 3) So, ๐๏ทฏ = 1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ Let ๐๏ทฏ = ๐๏ทฎ1๏ทฏ ๐๏ทฏ + ๐๏ทฎ2๏ทฏ ๐๏ทฏ + ๐๏ทฎ3๏ทฏ ๐๏ทฏ A line parallel to a plane is perpendicular to the normal of the plane. And two lines ๐๏ทฏ and ๐๏ทฏ are perpendicular if ๐๏ทฏ. ๐๏ทฏ = 0 Given, the line is parallel to planes So, our equations are ๐1 โ ๐2 + 2๐3 = 0 3๐1 + ๐2 + ๐3 = 0 Thus, ๐๏ทฏ = ๐๏ทฎ1๏ทฏ ๐๏ทฏ + ๐๏ทฎ2๏ทฏ ๐๏ทฏ + ๐๏ทฎ3๏ทฏ ๐๏ทฏ = โ3k ๐๏ทฏ + 5k ๐๏ทฏ + 4k ๐๏ทฏ Now, Putting value of ๐๏ทฏ & ๐๏ทฏ in formula ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ โด ๐๏ทฏ = (1 ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3k ๐๏ทฏ + 5k ๐๏ทฏ + 4k ๐๏ทฏ) = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐k (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) = ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 ๐๏ทฏ) Therefore, the equation of the line is ( ๐๏ทฏ + 2 ๐๏ทฏ + 3 ๐๏ทฏ) + ๐ (โ3 ๐๏ทฏ + 5 ๐๏ทฏ + 4 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.