calculus question?

Question

Im trying to find the parametric and symmetric equation for the line of the intersection between the two planes: 3x-2y+z=6 and z=4x+1/2y+5

Can someone assist me in solving the equation? I need something to get me started to solve the problem. Any help will be appreciated. thanks

Answer 1

To get the parametric equation of the line we have to set one of the 3 variable as the parameter. The easiest and obvious choice is z.

Let z = t (usual designation for parameter)

We try to express equation 1 in terms of x = f(y,t) - I'm not gonna show you the steps.

y= 1/2t + 1.5x - 3 ------ eq. (3)

From equation 2, express x = f(t), since the y can be substituted by eq. (3). After some simplification,

x = 3/19t - 14/19 ----- this is your first parametric eqn. (4)

Getting back to (3), substitute (4) into (3), so we now have y = f(t)...

y = 14/19t - 78/19

Therefore your parametric equations are:
z = t
x = 3/19t - 14/19
y = 14/19t - 78/19

Answer 2

We can take z = t
then 3x - 2y = - t + 6
and 4x + 1/2y = t - 5
or 8x + y = 2t - 10

then y = 2t - 10 - 8x replaced in the 1st equation :

3x - 4t + 20 + 16x = - t + 6
19x = 3t - 14
x = (3t - 14)/19

y = 2t - 20 - 8(3t - 14) / 19 = (14t - 268)/19

and z = t

We can take also y = t or x = t