find the parametric equation of prejection of the line x=t, y=t, z=t on the plane -x+y-2z = 0?

Question

I have no idea what this means?

Answer 1

Clearly the point (1,1,1) lies on the line.

The normal to the plane is along -i + j -2k.

What are the equations of the line through (1,1,1) which is perpendicular to the plane?

(x - 1)/(-1) = (y - 1)/(1) = (z - 1)/(-2) = u (new parameter).

What are the coordinates of the foot of the perpendicular?

u is given by: -(1- u) + (1 + u) -2(1 - 2u) = 0, hence u = 1/3.

The line of projection is the line through (0,0,0) and (2/3,4/3,1/3).

x = (3/2)v; y = (3/4)v; z = 3v. [New parameter v.]

Answer 2

Find the parametric equation of the projection of the
line x = t, y = t, z = t on the plane -x + y - 2z = 0.
___________

If you were to drop a perpendicular line thru every point on the given line and see where the perpendicular intersects the plane, you would get the projection of the given line onto the plane. The projection itself will be another line. We need to find the equation of that line.

By inspection we see that the origin O(0, 0, 0) is a point both on the given line and in the plane. It is the point of intersection.

The normal vector n, of the plane can be taken from the coefficients of the variables x, y, and z.

n = <-1, 1, -2>

Again by inspection we can see that the point P(1, 1, 1) is also on the given line. (Actually, any point where x, y, and z are equal is on the given line.) Let's find its projection onto the plane by finding the intersection of the perpendicular line thru P and the given plane. The equation of the perpendicular is:

L(s) = P + sn
L(s) = <1, 1, 1> + s<-1, 1, -2>
where s is a scalar ranging over the real numbers

Plug in the values into the equation of the plane to solve for s at the point of intersection.

-x + y - 2z = 0
-1(1 - s) + 1(1 + s) - 2(1 - 2s) = 0
-2 + 6s = 0
s = 1/3

x = 1 - s = 1 - 1/3 = 2/3
y = 1 + s = 1 + 1/3 = 4/3
z = 1 - 2s = 1 - 2/3 = 1/3

The projection of P onto the plane is Q(2/3, 4/3, 1/3). Let's now use the points of the projection on the plane O and Q to calculate the directional vector v, of the line of projection is the plane.

v = Q - O = <2/3, 4/3, 1/3>

Any non-zero multiple of v is also a directional vector. Multiply by 3.

v = <2, 4, 1>

The equation of the line of projection is:

L(u) = O + uv
L(u) = u<2, 4, 1>

Put the equation of the line of projection in parametric form.

L(u)
x = 2u
y = 4u
z = u

Answer 3

i didnt do my own homework and im not gonna do yours