Answer:
- (x, y, z) = (3, 2, 1)
- a. no intersection; b. L = (4, -2, 0) +t(-2, 2, 1)
Step-by-step explanation:
You want the point where the given line intersects the plane, and the line of intersection of two planes, for two pairs of planes.
1. Point
We can write the equation of the line as a parametric equation, substitute for the variables x, y, z in the plane equation, find the value of the parameter, then find the point.
The parametric equation for the line can be written by letting the value of each ratio be equal to the parameter t.
t = x -2 = (y +3)/5 = (z -4)/-3
(x, y, z) = t(1, 5, -3) +(2, -3, 4)
x -2y +3z -2 = (t +2) -2(5t -3) +3(-3t +4) -2 = t(1-10-9) +(2 +6 +12 -2) = 0
-18t +18 = 0 ⇒ t = 1
(x, y, z) = (1, 5, -3) +(2, -3, 4) = (3, 2, 1)
The point of intersection is (3, 2, 1).
2a. Line 1
The normal vector to each plane is the coefficients of the variables:
π1 = (1, 2, 3)
π2 = (2, 4, 6) = 2(1, 2, 3)
These vectors have the same direction, so the planes are parallel.
There is no line of intersection.
2b. Line 2
Solving the two equations in two unknowns, we find ...
x +2z = 4
y -2z = -2
Using z = t, we can write the parameterized equation for the line as ...
(x, y, z) = (4 -2t, -2 +2t, t)
or ...
L = (4, -2, 0) +t(-2, 2, 1)
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Additional comment
Another way to find the line of intersection is to determine its direction from the cross product of the normal vectors. Then, a point on the line must be found. This generally involves solving the plane equations for some arbitrary value of one of the variables. It seems easier to make use of the matrix row-reduction capability of a calculator to do both at once.
