Angle between two planes: $ x+2y-6 $ and $ y=0 $

Question

Plane 1: $x + 2y - 6 = 0$

Plane 2: $y = 0$

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I can see that the normal vector of the first plane is {1, 2, 0} but what would be the the normal vector of the second one?

Answer 1

For the plane $ax+by+cz=d$ the normal is $(a, b, c)$. For the plane $y = 0$ we may write $ 0x+1y+0z=0$. Hence the normal is $\boxed{(0, 1, 0)}$.

Answer 2

Your first normal vector is not correct. The normal vector should be $(1, 2, 0)$

You might ask: Why we do not coins $6$? No matter what constant we take, those planes are parallel, and thus have the same normal vector.

In terms of the second equation, write it as $$0\cdot x + 1 \cdot y + 0 \cdot z = 0$$

Could you find its normal vector?