I am a novice in mathematics, and I have a question:
Suppose that I have 3 points in the space:
- (x,y,z) for these points are not known for me.
- given that I know the angles a, b and c (c.f. above image); for each of the three vectors connecting each two points, with respect to the horizontal plane (and also the distance between each two points), is it possible to calculate the angle between the plane defined by these points and the horizontal plane?
Thank you in advance for any response.
Here is what I can make out:
Suppose I take 3 points $P_1$,$P_2$ and $P_3$. Hence the direction cosines(according to the question) are known. Let me assume the direction cosines as follows $$P_1P_2 \equiv <p_1,q_1,r_1>$$ $$P_2P_3 \equiv <p_2,q_2,r_2>$$ $$P_3P_1 \equiv <p_3,q_3,r_3>$$ We know that direction cosines is the direction ratios of a unit vector. Since the lengths of the given vectors are given(assuming them to be $l_1,l_2,l_3$),if we multiply $l_1$ with the d.c.s of $P_1P_2$ we get the direction ratios of of $P_1P_2$. Similarly we can get the direction ratios of the the other 2 vectors.
The direction ratio of the x-y plane is $$(0,0,1)$$ We can find the direction ratio os the normal of the plane containing the three points by taking the cross product of any of the two vectors(say $\overrightarrow {P_1P_2}$ and $\overrightarrow {P_2P_3 }$). Finally taking the dot product of the direction ratio of normal with that of x-y plane will give you the desired angle)