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

Ex 11.2, 9 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.2, 9 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3


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 ⃗ = (3𝑖 ̂ − 2𝑗 ̂ − 5𝑘 ̂) + 𝜆 ["(3" 𝑖 ̂−"2" 𝑗 ̂+"6" 𝑘 ̂")" −"(3" 𝑖 ̂−"2" 𝑗 ̂ −"5" 𝑘 ̂")" ] = 3𝑖 ̂ − 2𝑗 ̂ − 5𝑘 ̂ + 𝜆 ["(3" −3")" 𝑖 ̂−"(2" −(−2))𝑗 ̂+(6−(−5))𝑘 ̂)] A (3, − 2, − 5) 𝑎 ⃗ = 3𝑖 ̂ − 2𝑗 ̂ − 5𝑘 ̂ B (3, − 2, 6) 𝑏 ⃗ = 3𝑖 ̂ − 2𝑗 ̂ + 6𝑘 ̂ = 3𝑖 ̂ − 2𝑗 ̂ − 5𝑘 ̂ + 𝜆 [0𝑖 ̂ + 0𝑗 ̂ + 11𝑘 ̂] = 3𝒊 ̂ − 2𝒋 ̂ − 5𝒌 ̂ + 𝜆 (11𝒌 ̂) Therefore, the vector equation is 𝑟 ⃗ = 3𝑖 ̂ − 2𝑗 ̂ − 5𝑘 ̂ + 𝜆 (11𝑘 ̂) 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.