# Parametric Equations

• Sep 3rd 2009, 06:17 PM
purplerain
Parametric Equations
Find parametric equations of the line of intersection of the two planes
-8 x - 2 y + 3 z = 9
and
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 = ....] .

DirectionVector :=

EquationsOfLine :=

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?
• Sep 4th 2009, 03:35 AM
red_dog
Let z=t and solve the system formed by the two equations in respect to x and y.
• Sep 4th 2009, 05:00 AM
Calculus26
you could in fact set z = 0 to find a pt on the line and as you suggested

take the cross product of the normals -8i -2j +3k and 5i + 4j +2k to find the vector parallel to the line of intersection
• Sep 4th 2009, 05:43 AM
HallsofIvy
Quote:

Originally Posted by purplerain
Find parametric equations of the line of intersection of the two planes
-8 x - 2 y + 3 z = 9
and
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 = ....] .

DirectionVector :=

EquationsOfLine :=

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?

I hate these things were you are told to just plug values into "Maple" or some other machine! What are you expected to learn from that? The equations of the planes are -8x- 2y+ 3z= 9 and 5x+ 4y+ 2z= -8. Multiply the first equation by 2 to get -16x- 4y+ 6z= 18 and add to the second to eliminate y: -11x+ 8z= 10. Then 8z= 11x+ 10 and z= (11/8)x+ 5/4. Putting that back into the first equation, -8x- 2y+ 3((11/8)x+ 9/4)= -8x- 2y+ (33/8)x+ 27/4= 9 or (-31/8)x- 2y= 9/4 so -2y= (31/8)x+ 9/4 and y= (-31/16)x- 9/8. In particular, if we take x= 16t, we have y= -31t- 9/8 and z= 22x+ 5/4.
• Sep 4th 2009, 05:43 AM
purplerain
Ok, when you mean vector parallel to the line of intersection, do you mean unit vector? I solved for x, y, and z and got x =-10/11, y =-19/22, and z = 3. Would the equation of the line be 5 (-10/11)t+4(-19/22)t+2(3t)?
• Sep 4th 2009, 06:17 AM
Calculus26
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
• Sep 4th 2009, 09:25 PM
purplerain
Thanks guys for the help, it cleared up a lot of confusion that i had.