Find the parametric equations for the line of intersection of planes:?

Question

Find the parametric equations for the line of intersection of planes:

z= x+y, 2x-5y-z=1

Is it possible to set any x,y,z point equal to 0?

For instance my book sets x=0 and they get the points (0, -1/6, -1/6) & get the resulting parametric equations x=6t, y=(-1/6)+t, z=(-1/6)+7t

but when I did it I set z=0 and got points (1/7, -1/7, 0). I would get different parametric equations. Is this wrong?

Answer 1

All you really need are 2 points on the line. So, set x = 0, and we get

z = y, - z = 1+5y

And therefore, 0=6y+1, y=-1/6, z = -1/6.

So, our first point is (0, -1/6, -1/6)

Now, set z = 0 and get the equations

x+y = 0
2x - 5y = 1

Which gives you

-7y = 1

y=-1/7, x = 1/7, z = 0

Our two points are, as you said, (1/7, -1/7, 0) and (0, -1/6, -1/6)

All we need are two points, and our line is therefore

x = 0+t/7
y = -1/6 + t/42
z = -1/6 +t/6

This may seem different, but remember that we can just set t=42s, and these become

x = 6s
y=-1/6 + s
z = -1/6 + 7s

The parameter (s or t) does not matter, because s and t can both range over the real numbers from negative to positive infinity.

=)

Answer 2

Find the parametric equations for the line of intersection of the planes:

z = x + y
2x - 5y - z = 1

Let's recast the equations of the planes.

x + y - z = 0
2x - 5y - z = 1

The cross product of the normal vectors of the two planes will be the directional vector v, of the line of intersection.

v = <1, 1, -1> X <2, -5, -1> = <-6, -1, -7>

Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1.

v = <6, 1, 7>

Now find a point on the line. It will be a point in both planes. Let y = 0 and solve for x and z.

x - z = 0
2x - z = 1

Subtract the first equation from the second.

x = 1

Plug back into the first equation and solve for z.

1 - z = 0
z = 1

So our point on the line is P(1, 0, 1).

The equation of the line of intersection is:

L(t) = P + tv
L(t) = <1, 0, 1> + t<6, 1, 7>
where t is a scalar ranging over the real numbers

Now put the equation of the line in parametric form.
L(t):
x = 1 + 6t
y = t
z = 1 + 7t
__________

Both points (0, -1/6, -1/6) and (1/7, -1/7, 0) are on the line as is my point of P(1, 0, 1).

In order for your parametric equation to be correct you need:

1) Any non-zero multiple of the directional vector of the line.
2) Any point on the line.

So any of the three points, and an infinite number of other points on the line, along with a directional vector of the line will give you correct parametric equations. Your point is fine.

Answer 3

Parametric Representation Of A Line

Answer 4

How To Find Parametric Equations

Answer 5

The xy airplane is the airplane for which z=0. So the intersection of this airplane with x+y+3z=0 is obviously x+y=0. So the line is exact via x+y=0 and z=0. (i won't be able to undergo in concepts if it somewhat is "parametric equation" format or no longer).