Level 7 · Ages 17–18
Read it again any time, including while you work through the exercises.
Leave the sheet of paper: welcome to space! To locate a point, two numbers are no longer enough, you need three: the x-coordinate, the y-coordinate and the z-coordinate (the height). A point is written A(x, y, z), and almost everything you know from the plane extends naturally: vector AB has coordinates (xB - xA, yB - yA, zB - zA), still arrival minus departure, and the dot product becomes xx' + yy' + zz'.
In space, a line is still defined by a point and a direction vector, as in the plane. The novelty is the plane: an infinite flat surface, like a table without edges. A plane can be described by a Cartesian equation ax + by + cz + d = 0. The triple (a, b, c) is not decorative: it is a normal vector of the plane, meaning perpendicular to every line lying in the plane.
This normal vector is the key to almost everything: two planes are parallel when their normal vectors are collinear; a line is parallel to a plane when its direction vector is orthogonal to the normal vector (zero dot product); a line is perpendicular to the plane when its direction vector is collinear with the normal vector.
Space holds one last surprise: two lines can be neither intersecting nor parallel! Think of a highway and the bridge passing over it: they never meet, yet they are not parallel. We say they are skew (non-coplanar). In space, always check whether your objects live in one plane before reasoning as on your sheet.
In space, A(1, 0, 2) and B(3, 4, 3). Find the coordinates of vector AB.
x = 3 - 1 = 2
Arrival minus departure for each coordinate, exactly as in the plane. First the x's.
y = 4 - 0 = 4
Then the y's: 4 minus 0.
z = 3 - 2 = 1
And the novelty of space: the z-coordinate. 3 minus 2 equals 1. Three coordinates for a space vector.
Result: AB(2, 4, 1).
Plane P has equation 2x - y + z = 5. Does point A(1, -1, 2) belong to P?
2(1) - (-1) + 2
I substitute A's coordinates into the left side. Mind the double sign: minus (-1) gives +1.
2 + 1 + 2 = 5
The computation gives 5, exactly the right side. The equation holds: A is in the plane. Bonus: I read the plane's normal vector from the coefficients: n(2, -1, 1).
Result: Yes, A belongs to plane P.