given plane x+y+2z=3 find the intersection with the line paralel to vector<1,1,1> and pases through (-1,0,0)

Answer 1

Given the plane x + y + 2z = 3, find the intersection of the plane with the line parallel to the vector v = <1,1,1> that passes through the point P(-1,0,0).

The equation of the line is:

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

The equation of the plane is:

x + y + 2z = 3

The equations of the line and plane will be equal at the point of intersection. Plug the values of the variables of the line into the equation of the plane.

(-1 + t) + t + 2t = 3
4t = 4
t = 1

Solve for the point of intersection Q.

x = -1 + t = -1 + 1 = 0
y = t = 1
z = t = 1

The point of intersection of the line and the plane is
Q(0, 1, 1).

Answer 2

Your line is x = -1 + t ; y = 0 + t ; z = 0 + t

Stick this into your plane equation.

(-1 + t) + (t) + 2(t) = 3
-1 + 4t = 3
4t = 4
t = 1

x = -1 + t ; y = 0 + t ; z = 0 + t
x = 0 ; y = 1 ; z = 1