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Why the outward unit normal to the plane $z=0$ is $-\vec{k}$

this question arised because i am solving to find total flux from the unit cube $0\le x,y,z\le 1$

So in the book the unit normal vector of $z=0$ ,$x=0$ and $y=0$ are taken as:$-\vec{k},-\vec{i},-\vec{j}$ respectively.

can i get some clarity how outward unit normal direction is decided?

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The cube is in the first octant. The plane $z=0$ represents the $XY$ plane, for which there are two possible normal unit vectors: $\hat k$ and $-\hat k$. In your case, $\hat k$ would be pointing towards the interior of the cube, so the outward unit normal vector is $-\hat k$.

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Normal to the surface of the plane $ax + by + cz = k$ is $(a,b,c)$ or any scalar multiple thereof.

Normal to $z= 0$ then is $(0,0,1)$

As for outward, you will need to look at the described solid to decide which vector points into an which way points outward from the solid. You can really just use your intuition.

If you want to be more systematic, find a point in the interior, say $(\frac 12,\frac 12, \frac 12)$ and which vector will take you from a point on the surface toward the interior... and which will take you away.

For the plane $z = 1, (0,0,1)$ points outward.

For the plane $z = 0, (0,0,-1)$ points outward.

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Considering that your problem is about the direction of unit vector that is to be considered, the answer is that you always have to take the outward drawn unit normal vector. In case you take $ \hat k$ as the unit normal vector, it takes you inside the cube but not outside.

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