Find an eqn for the plane with the line with parametric eqns...?

Question

Find an equation for the plane that contains the line with parametric equations x = −1+3t, y = 5+2t,
z = 2 − t and is parallel to the plane 2x − 4y + 2z = 9.

Answer 1

Hmm. I've never done a problem like this before. If I had to guess, I'd say that the equation you're looking for looks like 2x - 4y + 2z = A, where A is some number not equal to 9.

I suppose It would be helpful to find an equation with x and y, then another with y and z, and a third with x and z.

Let t=0
x=-1 , y=5 , and z=2.
let t=1
x=2 , y=7 , and z=1

The x,y relationship contains points (-1 , 5) and (2 , 7)
The y,z relationship contains points (5 , 2) and (7 , 1)
The x,z relationship contains points (-1 , 2) and (2 , 1)

The three equations, out of order, are
1.) y = 2/3*x + 17/3
3y = 2x +17
2x - 3y = -17
2.) z = -1/3*x + 5/3
3z = -x +5
x + 3z = 5
3.) z = -1/2*y + 9/2
2z = -y +9
y + 2z = 9

And from these three equations you can create a system to solve for the values of x, y, and z. The solution is...shoot. This system has an infinite number of solutions. Of course it does. Ah, well. I tried. Had the system had a solution, it would have been a simple enough matter of plugging the values for x, y, and z into the equation in the first paragraph, to find the value for A.

Sorry I couldn't help. Maybe you can see where I went wrong? Good luck.

Answer 2

Find an equation for the plane that contains the line with parametric equations

x = −1 + 3t
y = 5 + 2t
z = 2 − t

and is parallel to the plane 2x − 4y + 2z = 9.
____________

First check to see if the directional vector v, of the line lies in the plane.

v = <3, 2, -1>

The directional vector of the line should sum to zero in the plane if it lies in it.

2*3 − 4*2 + 2*(-1) = 6 - 8 - 2 = -4 ≠ 0

So the line cannot be in any plane parallel to the given plane.

Perhaps you meant for the parametric equation for z to be:
z = 2 + t

In that case the vector lies in the plane and we can proceed.
_____________

If the plane is parallel to the given plane, the x, y, and z coefficients will be the same. Only the constant will be different. Since the desired plane contains the given line, pick a point on the line and plug the point into the equation of the plane to find the constant. Select t = 0 which gives the point P(-1, 5, 2).

2x − 4y + 2z = 2*(-1) - 4*5 + 2*2 = -2 - 20 + 4 = -18

The equation of the desired plane is:

2x − 4y + 2z = -18

Divide by 2 to simplify.

x - 2y + z = -9

Answer 3

locate the equation of the line passing in the time of the factor B(a million,-3,0) and perpindicular to the plane x + 2y - z = 10. because of the fact the line is perpendicular to the plane, the traditional vector of the plane is likewise the directional vector of the line. and you have a factor B(a million, -3, 0). The directional vector v, of the line (and primary vector of the plane) is: v = Now write the equation of the line. L(t) = B + television L(t) = + t the place t is a scalar ranging over the authentic numbers