# Ex 11.2, 9

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 − 𝑥1𝑥2 − 𝑥1 = 𝑦 − 𝑦1𝑦2 − 𝑦1 = 𝑧 − 𝑧1𝑧2 − 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 − 33 − 3 = 𝑦 − ( − 2) − 2 − ( − 2) = 𝑧 − ( − 5)6 − ( − 5) 𝒙 − 𝟑𝟎 = 𝒚 + 𝟐𝟎 = 𝒛 + 𝟓𝟏𝟏

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .