Exams › GATE › Engineering Mathematics

The unit normal vector to the surface \(x^2 + y^2 + z^2 - 48 = 0\) at the point \((4,4,4)\) is:

1. 1/√2, 1/√2, 1/√2
2. 1, 1, 1
3. -1/√2, -1/√2, -1/√2
4. -1, -1, -1

Correct answer: 1/√2, 1/√2, 1/√2

Solution

For the surface \(F(x,y,z)=x^2+y^2+z^2-48=0\), the normal vector is given by \(\nabla F=(2x,2y,2z)\). At \((4,4,4)\), this is \((8,8,8)\), whose unit vector is \((1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})\); the provided option set appears OCR-corrupted, but the keyed option is the first one.

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