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Ex 11.2, 4 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Ex 11.2
Ex 11.2, 1
Ex 11.2, 2
Ex 11.2, 3 Important
Ex 11.2, 4 You are here
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Ex 11.2, 7 Important
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Ex 11.2, 9 Important
Ex 11.2, 10 (i) Important Deleted for CBSE Board 2022 Exams
Ex 11.2, 10 (ii)
Ex 11.2, 11 (i) Important Deleted for CBSE Board 2022 Exams
Ex 11.2, 11 (ii)
Ex 11.2, 12 Important
Ex 11.2, 13
Ex 11.2, 14 Important
Ex 11.2, 15 Important
Ex 11.2, 16
Ex 11.2, 17 Important
Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Serial order wise
Transcript
Ex 11.2, 4 Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3 𝑖 + 2 𝑗 – 2 𝑘 . Equation of a line passing through a point with position vector 𝑎 , and parallel to a vector 𝑏 is 𝑟 = 𝑎 + 𝜆 𝑏 Since line passes through (1, 2, 3) 𝑎 = 1 𝑖 + 2 𝑗 + 3 𝑘 Since line is parallel to 3 𝑖 + 2 𝑗 − 2 𝑘 𝑏 = 3 𝑖 + 2 𝑗 − 2 𝑘 Equation of line 𝑟 = 𝑎 + 𝜆 𝑏 𝒓 = ( 𝒊 + 2 𝒋 + 3 𝒌) + 𝜆 (3 𝒊 + 2 𝒋 − 2 𝒌)
