Vectors, cross products, lines and planes

Try

Put a point P at (1,1,0) and another point Q at (3,2,2). The displacement from P to Q is (2,1,2), whose length is 3. Coordinates locate a point relative to an origin; a displacement compares two points and does not depend on where that origin was placed. Draw the horizontal displacement (2,1,0) first, then the vertical displacement (0,0,2). A perspective sketch helps organize components, but its screen lengths are not the actual spatial lengths. All coordinates in this example use the same length unit.

Predict

Predict which of two directions, a=(1,0,1) and b=(0,2,0), is perpendicular to the displacement Q−P. Compute dot products before deciding: a·(Q−P)=4 and b·(Q−P)=2, so neither is perpendicular. Now construct a perpendicular direction instead of guessing. The vector (1,−2,0) has dot product zero. Explain why multiplying this direction by any nonzero scalar preserves perpendicularity. A normal vector specifies a direction, not a unique length.

Experiment

Construct the line r(t)=P+t(Q−P), where t is a real parameter. At t=0 you are at P; at t=1 you are at Q; allowing every real t extends beyond the segment. Intersect it with the plane z=1 by solving 2t=1. The result is (2,3/2,1). Check all three coordinates in the line and the plane, rather than testing only the coordinate that determined t.

Parameter t	x	y	z
0	1	1	0
1/2	2	3/2	1
1	3	2	2

The table is an exact coordinate trace, not a projection drawing.

Understand

For vectors a=(a₁,a₂,a₃) and b=(b₁,b₂,b₃), their dot product is a₁b₁+a₂b₂+a₃b₃. It also equals |a||b|cos θ for nonzero vectors. Their cross product is (a₂b₃−a₃b₂, a₃b₁−a₁b₃, a₁b₂−a₂b₁). It is perpendicular to both; its magnitude is their parallelogram area. Order matters: b×a=−a×b.

For u=(1,1,0) and v=(0,1,2), u×v=(2,−2,1). The plane through the origin spanned by u,v is therefore 2x−2y+z=0. Substitution verifies both spanning directions. The scalar triple product a·(b×c) is a signed volume; its absolute value is the parallelepiped volume. A zero triple product means the three directions are linearly dependent.

Apply

Repair this argument: “Two lines have different direction vectors, so they intersect.” In three dimensions, nonparallel lines can be skew. To test intersection, equate all three coordinate expressions using separate parameters and require one pair of parameter values to satisfy all equations. For example, (t,0,0) and (0,s,1) never intersect because their z-coordinates differ.

For a line p+td and plane n·(x−q)=0, substitution produces n·(p−q)+t n·d=0. A nonzero n·d gives exactly one intersection. If it is zero, the line is either entirely in the plane or disjoint from it; inspect the constant term to distinguish these outcomes.

Continue

As optional depth, relate cross products to determinants and explain why cross-product coordinates change sign when two input vectors are swapped. The geometric area does not change sign, but the chosen normal does. Later surface integrals will use precisely this orientation information.

Keep a record containing a spatial sketch, a coordinate calculation and a verbal explanation of one intersection. These representations catch different errors. The vector span investigation offers a lower-dimensional experiment with combinations of directions; its plane picture does not replace checking three-dimensional consistency. Next, let the position depend nonlinearly on a parameter and ask how motion changes along the resulting curve.