# angle between steepest gradient of two plane

IF I have two 3d planes such as Oab and Oa'b'. If these two planes intersect a horizontal plane and the intersection of each plane makes AB and A'B' lines. then,

1. Does the angle between AB, A'B' i.e. APA' is equal to the angle between XY projected normal vectors (i.e. XY projected n1 and n2 or angle between steepest gradient) of the planes?

2. Does this scenario is always true, even if we intersect these two planes with third plane which is not horizotal (i.e. instead of XY , if there is oblique plane)?

Please, answer my both questions. Regarding your first question, WLOG you can move your normal vectors to intersect with $AB$ and $A'B'$ respectively. Then, consider the set of all the planes that would generate the same line $AB$ (it is a rotation of $Oab$ along $AB$ axis). There is a bijection between normal vectors and such planes (because of the aforementioned rotation), moreover, the set of endpoints of normal vectors forms a circle which plane is perpendicular to $AB$. Therefore, any projection of $n_1$ into $XY$ will be perpendicular to $AB$ (or equal to zero vector). Following similar reasoning for the second plane we got that, indeed, if the $XY$ projected normal vectors $n_1$ and $n_2$ are non-zero, then the angle between them is the same as the angle between $AB$ and $A'B'$.

As for your second question, you can always introduce a new (local) coordinate system in which the $XY$ plane will be the $\{z = 0\}$ plane.

I hope this helps ;-)

• thanks for the explanation, Instead of a horizontal plane, if I take another oblique plane to the horizon, then does the above angle case also same?
– gnp
Mar 28, 2013 at 11:27
• @gnp There is no horizontal plane in $\mathbb{R}^3$, every direction, every plane is the same. There is the horizontal plane when you introduce coordinates, but those are not required for the angles to be the same. Mar 28, 2013 at 11:39
• ok, got thanks..
– gnp
Mar 28, 2013 at 11:40