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Ex 11.3, 2 - Find vector equation of a plane which is 7 units - Equation of plane - In Normal Form

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.3, 2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3 𝑖﷯ + 5 𝑗﷯ − 6 𝑘﷯. Vector equation of a place at a distance ‘d’ from the origin and normal to the vector 𝑛﷯ is 𝒓﷯ . 𝒏﷯ = d Unit vector of 𝑛﷯ = 𝑛﷯ = 1﷮ 𝑛﷯﷯﷯( 𝑛﷯) Distance form origin = d = 7 𝑛﷯ = 3 𝑖﷯ + 5 𝑗﷯ − 6 𝑘﷯ Magnitude of 𝑛﷯ = ﷮32+52+ − 6﷯2﷯ 𝑛﷯﷯ = ﷮9+25+36﷯ = ﷮70﷯ Now, 𝑛﷯ = 1﷮ 𝑛﷯﷯﷯( 𝑛﷯) = 1﷮ ﷮70﷯﷯(3 𝑖﷯ + 5 𝑗﷯ − 6 𝑘﷯) = 3 𝑖﷯ + 5 𝑗﷯ − 6 𝑘﷯﷮ ﷮70﷯﷯ Vector equation is 𝑟﷯. 𝑛﷯ = d 𝒓﷯. 𝟑 𝒊﷯ + 𝟓 𝒋﷯ − 𝟔 𝒌﷯﷮ ﷮𝟕𝟎﷯﷯﷯ = 7 Therefore, the vector equation of the plane is 𝑟﷯. 3 𝑖﷯ + 5 𝑗﷯ − 6 𝑘﷯﷮ ﷮70﷯﷯﷯ = 7

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