Left Hand Wall
Parameterizing this
where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y component is in the 4th quadrant.
(x y z
)=(cos 6\[Degree] -sin 6\[Degree] 0 sin 6\[Degree] cos 6\[Degree] 0 0 0 1
) . (x' y' z'
)
(x y z
)=(cos 6\[Degree] -sin 6\[Degree] 0 sin 6\[Degree] cos 6\[Degree] 0 0 0 1
) . (t cos (29.5\[Degree])+0.09156 -t sin (29.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos 6 \[Degree]cos (29.5\[Degree])+t sin 6 \[Degree]sin (29.5\[Degree]) -t cos 6 \[Degree]sin (29.5\[Degree])+0.09156 sin 6 \[Degree]+t cos (29.5\[Degree])sin 6 \[Degree] 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos(29.5\[Degree])+ sin 6 \[Degree]sin (29.5\[Degree])) 0.09156 sin 6 \[Degree]-t (cos 6 \[Degree]sin (29.5\[Degree])-sin 6 \[Degree] cos (29.5\[Degree])) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (6\[Degree] -29.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (6 \[Degree]-29.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (-23.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (-23.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (23.5\[Degree]) 0.09156 sin 6 \[Degree]-t sin (-23.5\[Degree]) 0
)
Using the equation for y we can solve for t
Substituting this into the expression for x