Find the point at which the line intersects the given plane.
x = 1 + 4t, y = 4t, z = 2−3t ;
x + 2y − z + 1 = 0
(x, y, z) = _____________?
(1+4t) +2 (4t) -(2-3t) = -1
solve for t
then use that t to find x, y z
To find the point at which the line intersects the given plane, we need to substitute the expressions for x, y, and z from the line equation into the plane equation and solve for t.
The plane equation is x + 2y - z + 1 = 0.
Substituting the expressions for x, y, and z:
(1 + 4t) + 2(4t) - (2 - 3t) + 1 = 0
Simplifying the equation:
1 + 4t + 8t - 2 + 3t + 1 = 0
15t = 0
t = 0
Now that we have the value of t, we can substitute it back into the expressions for x, y, and z to find the point of intersection.
x = 1 + 4(0) = 1
y = 4(0) = 0
z = 2 - 3(0) = 2
Therefore, the point of intersection is (1, 0, 2).
To find the point at which the line intersects the given plane, we need to find the values of x, y, and z that satisfy both the equations of the line and the equation of the plane.
First, let's write the equation of the line in terms of t:
x = 1 + 4t
y = 4t
z = 2 - 3t
Now, substitute these expressions for x, y, and z into the equation of the plane:
(1 + 4t) + 2(4t) - (2 - 3t) + 1 = 0
Simplifying the equation:
1 + 4t + 8t - 2 + 3t + 1 = 0
15t = 0
t = 0
Now that we have the value of t, we can substitute it back into the equations of the line to find the values of x, y, and z.
Substituting t = 0 into the equations of the line:
x = 1 + 4(0) = 1
y = 4(0) = 0
z = 2 - 3(0) = 2
Therefore, the point of intersection of the line and the plane is (1, 0, 2).