Thread: Normal to plane.

I know the plane equation 0.00023x - 0.00018y + 0.00032z + 1 = 0
I need to calculate the angle between the normal of this plane and horizontal (x-y plane).
Please help me to calculate that.
Could you suggest any tutorials about this

Hi, swee.

You could write the equation of your plane as

0.00023x-0.00018y+0.00032z=-1

This is now in the form ax+by+cz=d. The reason this is helpful is because a normal to the plane written in this form is <a,b,c>. So, you could take your normal vector to be <0.00023, -0.00018, 0.00032>, but the decimals are a bit annoying. So we can scale up by a factor of 10^5 to get the vector <23, -18, 32> that's still normal to the plane and will have the same angle with the xy-plane as <0.00023, -0.00018, 0.00032>, because <23, -18, 32> is parallel to <0.00023, -0.00018, 0.00032>.

Does this help you get started? Good luck!

thanks a lot. Now i understood how to get normal to plane. could you please help me in calculating angles between this normal vector <23, -18, 32> and horizontal ?

angle can be calculated by getting dot product. but how can we represent horizontal (x-y plane) in vector form?

is it ok if i represent horizontal vector as <0,0, 1> and get dot product to both vectors

Well, you don't want to use <0,0,1> as your horizontal vector, because <0,0,1> is a unit vector in the vertical direction.

Your idea to use the dot product is a good one. What we want to do is determine the projection of <23, -18, 32> onto the xy-plane and take a dot product with this projection. If we write

<23, -18, 32> = <23, -18, 0> + <0, 0, 32>

we see that the projection onto the xy-plane is <23, -18, 0>.

Does this approach make sense? Good luck!