Vector question....hmmmm??

Question

The line L through the origin is normal to the plane 2 - y - z = 4. Find the point in which L meets the plane x + y - 2z = 2....

Answer 1

I assume you mean the first plane is:
2x - y - z = 4

Since the line is perpendicular to the plane above, the directional vector v, of the line is the same as the normal vector to the plane.

L(t) = O + tv
L(t) = <0, 0, 0> + t<2, -1, -1>
L(t) = t<2, -1, -1>

Find the point P where the line L meet the plane:
x + y - 2z = 2.

Plug in the values of the the variables x, y, and z from the line into the equation of the plane.

1(2t) + 1(-t) - 2(-t) = 2
2t - t + 2t = 2
3t = 2
t = 2/3

The point P of intersection is:
x = 2t = 2(2/3) = 4/3
y = -t = -1(2/3) = -2/3
z = -t = -1(2/3) = -2/3

The point of intersection is:
P(4/3, -2/3, -2/3)

Answer 2

I hope you mean the first plane is 2x - y - z = 4
so that the d.r. of the normal are 2 : -1 : -1

therefore normal to this plane from the origin has the form

[x/2] =[ y/ (-1) ] = [ z/( -1) ] = t

or parametric form is
x= 2t
y = -t
z = -t

just substitute into the equation
x + y - 2z = 2

solve for t and then for x, y, z from the parametric equation of the line.