1. Considering 2 planes with equations:
x + 2y - 3z = 1
x + 2y - 3z = 18
(a) Given (a,b,c) a point on the first plane, find the point where the line perpendicular to both planes passes by (a,b,c) and through the 2nd plane.
To find the point where the line perpendicular to both planes passes through a given point on the first plane and the second plane, we need to determine the direction vector of the line and then find the intersection point.
First, let's find the normal vectors of both planes. The normal vector of a plane with equation Ax + By + Cz = D is (A, B, C).
So, the normal vector of the first plane is (1, 2, -3) and the normal vector of the second plane is also (1, 2, -3) since both planes have the same equation x + 2y - 3z.
To find the direction vector of the line perpendicular to both planes, we can take the cross product of the normal vectors.
Let's call this direction vector d.
d = (1, 2, -3) x (1, 2, -3)
= (2*(-3) - 2*(-3), -3*1 - 1*(-3), 1*2 - 2*1)
= (-6 + 6, -3 + 3, 2 - 2)
= (0, 0, 0)
The resulting direction vector is (0, 0, 0), which means the line is parallel to both planes. In this case, there are infinitely many points where the line passes through both planes, including the given point (a, b, c) on the first plane.
Therefore, the point where the line passes through the given point (a, b, c) and the second plane is simply the given point itself: (a, b, c).