The vector that is normal to the surface 2x z2 – 3x y – 4 x = 7 at the point (1, –1, 2) is

Q. The vector that is normal to the surface 2x z2 – 3x y – 4 x = 7 at the point (1, –1, 2) is

            (A) 2i – 3j + 8k                         (B) 2i + 3j + 4k

            (C) 7i – 3j + 8k                           (D) 7i – 5j + 8k

Ans: 7i – 3j + 8k

Sol:

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