Given: -3x+9y=51 ···(1), 7x-6y=-16 ···(2) and z=x-y ···(3)
Draw imaginary three-dimensional space by Cartesian coordinates where each
of three axes, x, y and z, meets perpendicularly with the other two at the origin,
O(0,0,0), and x-y plane is horizontal, and y-z and z-x planes are vertical.
Let a point on x-y plane be Q(x,y,0) where line (1) and (2) intercept each other.
Let the other point be P(x,y,z) vertically z units apart from point Q, so point P
lies on an axis parallel to z-axis and perpendicular to x-y plane.
Here, we try to find the location of terminal point of vector OP: P(x,y,z).
From (1) and (2), we have: x=18/5 and y=103/15 (calculations are skipped),
so the coordinates of point Q is: (x,y,z)=(18/5,103/15,0)
From (3) and the results obtained above, we have: z=-49/15,
so the coordinates of point P is: ( x,y,z)=(18/5,103/15.-49/15)
Let the direction angle between x-axis and vector OQ , and the one between z-axis and vector OP be α and β respectively.
So, tanα=y/x=(103/15)/(18/5)=103/54 (α=arctan(103/54)≈62.33°)
tanβ=√(x²+y²) / z=√{(18/5)²+(103/15)²} / (-49/15)=-(5√541)/49≈-2.37
(β≈arctan(-2.37)≈247.12°(=67.12°+180°))
The answers are: (x,y,z)=(18/5,103/15,-49/15), tanα=103/54 and tanβ=(-5√541)/49