Find all points of intersection of the line {p:p = t(1, -3, 6)} and the plane {p:x + 3y + z = 2}
To find the points of intersection between the line and the plane, we need to solve the system of equations formed by the equation of the line and the equation of the plane.
The equation of the line is given by:
p = t(1, -3, 6) = (t, -3t, 6t)
The equation of the plane is given by:
x + 3y + z = 2
Substituting the parametric equations of the line into the equation of the plane, we have:
t + 3(-3t) + 6t = 2
Simplifying, we get:
t - 9t + 6t = 2
-2t = 2
t = -1
Substituting t = -1 back into the parametric equations of the line, we get:
p = (-1, 3, -6)
Therefore, the point of intersection between the line and the plane is (-1, 3, -6).
To find the points of intersection between a line and a plane, we need to solve the system of equations formed by the line and the plane.
The equation of the line is given as p = t(1, -3, 6), and the equation of the plane is given as x + 3y + z = 2.
We can substitute the parametric form of the line equation into the plane equation to find the points of intersection. Substituting p = t(1, -3, 6) into x + 3y + z = 2, we get:
t(1) + 3t(-3) + t(6) = 2
This simplifies to:
t - 9t + 6t = 2
-3t = 2
t = -2/3
Now we can substitute this value of t back into the line equation to find the corresponding point:
p = (-2/3)(1, -3, 6)
p = (-2/3, 2, -4)
Therefore, the line {p = t(1, -3, 6)} intersects the plane {x + 3y + z = 2} at the point (-2/3, 2, -4).