1. Class 12
2. Important Question for exams Class 12

Transcript

Ex 11.2, 5 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 ๐๏ทฏ โ ๐๏ทฏ + 4 ๐๏ทฏ and is in the direction ๐๏ทฏ + 2 ๐๏ทฏ โ ๐๏ทฏ . Equation of a line passing though a point with position vector ๐๏ทฏ and parallel to vector ๐๏ทฏ is ๐๏ทฏ = ๐๏ทฏ + ๐ ๐๏ทฏ Here, ๐๏ทฏ = 2 ๐๏ทฏ โ ๐๏ทฏ + 4 ๐๏ทฏ & ๐๏ทฏ = ๐๏ทฏ + 2 ๐๏ทฏ โ ๐๏ทฏ So, ๐๏ทฏ = (2 ๐๏ทฏ โ ๐๏ทฏ + 4 ๐๏ทฏ) + ๐ ( ๐๏ทฏ + 2 ๐๏ทฏ โ ๐๏ทฏ) โด Equation of line in vector form is (2 ๐๏ทฏ โ ๐๏ทฏ + 4 ๐๏ทฏ) + ๐ ( ๐๏ทฏ + 2 ๐๏ทฏ โ ๐๏ทฏ) Equation of a line passing though (x1, y1, z1) and parallel to a line having direction ratios a, b, c is ๐ฅ โ ๐ฅ1๏ทฎ๐๏ทฏ = ๐ฆ โ ๐ฆ1๏ทฎ๐๏ทฏ = ๐ง โ ๐ง1๏ทฎ๐๏ทฏ Since the line passes through a point with position vector 2 ๐๏ทฏ โ ๐๏ทฏ + 4 ๐๏ทฏ, โด ๐ฅ1 = 2, y1 = โ 1 z1 = 4 Also, line is in the direction of ๐๏ทฏ + 2 ๐๏ทฏ โ ๐๏ทฏ, Direction ratios : ๐ = 1, b = 2 , c = โ1 Equation of line in Cartesian form is ๐ฅ โ 2๏ทฎ1๏ทฏ = ๐ฆ โ ( โ 1)๏ทฎ2๏ทฏ = ๐ง โ 4๏ทฎ โ 1๏ทฏ ๐ โ ๐๏ทฎ๐๏ทฏ = ๐ + ๐๏ทฎ๐๏ทฏ = ๐ โ ๐๏ทฎ โ ๐๏ทฏ

Class 12
Important Question for exams Class 12