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A surface is given by 𝑧^2= 2đ‘Ĩ^2− đ‘Ļ^2 and 𝑛⃗ and −𝑛⃗ are unit normal vectors to the surface ...

 

Given, đ‘§^2= 2đ‘Ĩ^2− đ‘Ļ^2

F(x, y, z) = 2đ‘Ĩ^2− đ‘Ļ^2 - 𝑧^2

normal vector, ∇F = (4x, -2y, -2z)

Evaluating at P = (1,0,√2)

∇F(1,0,√2) = (4, 0, 2√2)

Normalizing the vector we get
√2 i/ √3 - 1k/√3

Answer D


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