Let z=t and solve the system formed by the two equations in respect to x and y.
Find parametric equations of the line of intersection of the two planes
-8 x - 2 y + 3 z = 9
5 x + 4 y + 2 z = -8
Assign your direction vector, as a Maple list, to the name DirectionVector. Use t as the parameter along the line and assign the equations of the line to the list EquationsOfLine in the form [x=.... , y = ..... , z = ....] .
How do i start this? Should i take the cross product and then use the results as the answer, or should i solve for one equation and substitute into the other?
When you take the cross product of -8i -2j +3k and 5i + 4j +2k
you obtain -16 i + 31 j - 22k
this is the direction vector of the line since it is perpindicular to both planes it is parallel to the line of intersection --not a unit vector
In letting z = 0 you obtain x = -10/11 and y = -19/22
So (-10/11,-19/22,0) is a point on the line
We have then x = -10/11 -16t
y= -19/22 +31t
z = -22t
Obviously thre is no unique answer