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Ex 11.2, 9 - Find vector and Cartesian equations of line - Ex 11.2

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.2, 9 Find the vector and the Cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6). Vector Equation Vector equation of a line passing through two points with position vectors 𝑎﷯ and 𝑏﷯ is 𝑟﷯ = 𝑎﷯ + 𝜆 ( 𝑏﷯ − 𝑎﷯) Given, the two points are So, 𝑟﷯ = (3 𝑖﷯ − 2 𝑗﷯ − 5 𝑘﷯) + 𝜆 (3 𝑖﷯−2 𝑗﷯+6 𝑘﷯)−(3 𝑖﷯ −2 𝑗﷯ −5 𝑘﷯)﷯ = 3 𝑖﷯ − 2 𝑗﷯ − 5 𝑘﷯ + 𝜆 (3− 3) 𝑖﷯−(2−(−2)) 𝑗﷯ + (6−(− 5)) 𝑘﷯)﷯ = 3 𝑖﷯ − 2 𝑗﷯ − 5 𝑘﷯ + 𝜆 [0 𝑖﷯ + 0 𝑗﷯ + 11 𝑘﷯] = 3 𝒊﷯ − 2 𝒋﷯ − 5 𝒌﷯ + 𝜆 (11 𝒌﷯) Therefore, the vector equation is 𝑟﷯ = 3 𝑖﷯ − 2 𝑗﷯ − 5 𝑘﷯ + 𝜆 (11 𝑘﷯) Cartesian equation Cartesian equation of a line passing through two points A(x1, y1, z1) and B (x2, y2, z2) is 𝑥 − 𝑥1﷮𝑥2 − 𝑥﷮1﷯﷯ = 𝑦 − 𝑦1﷮𝑦2 − 𝑦1﷯ = 𝑧 − 𝑧1﷮𝑧2 − 𝑧1﷯ Since the line passes through A (3, −2, −5) x1 = 3, y1 = − 2, z1 = − 5 And also passes through B (3, − 2, 6) x2 = 3, y2 = − 2, z2 = 6 Equation of line is 𝑥 − 3﷮3 − 3﷯ = 𝑦 − ( − 2)﷮ − 2 − ( − 2)﷯ = 𝑧 − ( − 5)﷮6 − ( − 5)﷯ 𝒙 − 𝟑﷮𝟎﷯ = 𝒚 + 𝟐﷮𝟎﷯ = 𝒛 + 𝟓﷮𝟏𝟏﷯

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