Question

Find a set of parametric equations for the line of intersection of the planes 3x-2y-z=7 x-4y+2z=0

Answer 1

Answer:

equation of the line is

L(x,y,z) =  (0,-7/4,-7/2) +(2,7/4, 5/2)*t , where t is the parameter

Step-by-step explanation:

for the planes

3x-2y-z=7

x-4y+2z=0

multiplying by 2 the first equation and summing both equations

7x-8y=14

for y=0 , x=2 →  z= -1 (replacing in any plane equation)

for x=0 , y=-7/4 →  z=-7/2 (replacing in any plane equation)

then (2,0,-1) and (0,-7/4,-7/2) belongs to the line → thus one parallel vector to the line is

v= (2,0,-1) -  (0,-7/4,-7/2) = (2,7/4, 5/2)

the parametric equation of the line is

L(x,y,z) =  (0,-7/4,-7/2) +(2,7/4, 5/2)*t , where t is the parameter

Note

for t=0 → L(x,y,z)= (0,-7/4,-7/2)

for t=1 →  L(x,y,z)= (2,0,-1)

Answer 2

Answer:

To find a set of parametric equations for the line of intersection of the planes 3x+2y-z=7        eqn1

x-4y+2z=0        eqn2

Mulitply eqn 1 by 2, to solve this simultaneously

6x+4y-2z=14       eqn3

 x -4y+2z=0

Add eqn 3 to eqn 2, we have

7x = 14

x = 2, put this into eqn2 and assume y = v

2 - 4v +2z =0

2z = 4v-2,     z=2v-1

Hence, a set of parametric equations are x = 2, y = v and z=2v-1