Suppose z = f(x, y), f(1, 3)= 5 and $\nabla f ( 1,3 ) = 4 \vec { i } + 5 \vec { j }$ the vector $- 4 \vec { i } + 5 \vec { j } + \vec { k }$ is perpendicular to the graph of f(x, y)at the point (1, 3).

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The gradient vector $\nabla f ( 1,3 ) = 4 \vec { i } + 5 \vec { j }$ is perpendicular to the level curve of $f$ at the point $(1, 3)$, not the vector $- 4 \vec { i } + 5 \vec { j } + \vec { k }$. The correct normal vector to the surface at a point includes the gradient and the change in the function itself, which would be represented by $4 \vec { i } + 5 \vec { j } - \vec { k }$ if considering the z-component as the change in $f$.

Suppose Z = F(x, Y), F(1, 3)= 5 And ∇f(1,3)=4i⃗+5j⃗\nabla f ( 1,3 ) = 4 \vec { i } + 5 \vec { j }∇f(1,3)=4i+5j​

Suppose Z = F(x, Y), F(1, 3)= 5 And $\nabla f ( 1,3 ) = 4 \vec { i } + 5 \vec { j }$