determine the line of intersection (in parametric form) of the following planes.

Question

The direction vector within you line must be stated without fractions

Line1: 3x -y + 2z = 4

Line2: 6x + 2y - z = 3

Answer 1

plane1 is:3x -y + 2z = 4

plane2 is 6x + 2y - z = 3

solving equations by eliminating x variable

y=5z/4-5/4

substituting y into first equation

x= -z/4+11/12

let z=4t

then y=5t-5/4

and x=-t+11/12

is parametic form of line

let's check something. If t=0.z=0 then x=11/12 and y= -5/4. Are those points on both planes?

3*11/12-(-5/4)=4, good

6*11/12+2*(-5/4)=3 good

Answer 2

Take the coefficients and calculate the cross-product to get the direction of the line.

i j k

3 -1 2

6 2 -1

-3i + 15j + 12k

Divide by 3

We get

-i + 5j + 4k

The direction of the line will be (-1, 5, 4)

Find a point of intersection, say, by setting z = 0

3x - y = 4

6x + 2y = 3

Double the top equation and add

12x = 11

x = 11/12

3(11/12) - y = 4

y = - 5/4

Thus, our line starting point is (11/12, -5/4, 0) and the direction is (-1, 5, 4)

Our solution is (11/12 - t, -5/4 + 5t, 4t)

As a check, plug this into the left side of the equation of each of the planes and see that we get the right side.