Question

Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x+3y+2z+2=0 2x−3y+2z−2=0

Answer 1

The solution is:

x = - 4×t

y = -4/6 + 4/6 t for 0 ≤ t ≤ 1

• First, we need to find the equation of the intersection line.

To look for the intersection line, we have:

3×x + 3×y + 2×z + 2 = 0 or 2×z = - 3×x - 3×y - 2 (1)

2×x - 3×y + 2×z -2 = 0 or 2×z = - 2×x + 3×y + 2 (2)

Solving the two equation sysstem in terms of x and y. Equalitying (1) and (2)

- 3×x - 3×y - 2 = - 2×x + 3×y + 2

reordering - x - 6×y - 4 = 0

So we got the equation of the intersection line

If in that equation we look for the intersection points, we get:

x = 0 - 6×y - 4 = 0 y = -4/6 ( intersection point P with y-axis)

y = 0 -x - 4 = 0 x = -4 ( intersection point Q with x-axis)

Now that we got the whole information about the intersection line ( see attached drawing)

• we proceed to parametrize the intersection line

Looking at the attached file we can see that the segment PQ is

PQ = ( -4 ; 4/6) ( coordinates of Q minus coordinates of P )

We choose t as a parameter then:

When t = 0 x = 0 x = -4×t x = 0

When t = 1 x = -4 x = -4×t x = -4

Now for y

When t = 0 y = - 4/6 y = -4/6 + 4/6 t y = -4/6

When t = 1 y = 0 y = -4/6 + 4/6(1) y = 0

finally x = - 4×t

y = -4/6 + 4/6 t for 0 ≤ t ≤ 1

Answer 2

The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).

Using x=0, we can find a point on this line by solving the simultaneous equations that remain:

... 3y +2z = -2

... -3y +2z = 2

Adding these, we get

... 4z = 0

... z = 0

so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...

• x = 12t
• y = -2/3 -2t
• z = -15t

By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).

Then our parametric equations can be written as

• x = 8 +12t
• y = -2 -2t
• z = -10 -15t