1.6 Vectors

A plane through point a with normal vector n has equation n·(r−a)=0, or equivalently n·r=d. The normal is perpendicular to every direction lying in the plane.

To find a normal, use a cross product of two non-parallel direction vectors in the plane. Check a candidate point by substitution and keep vector notation consistent.

A plane through (1,0,2) with normal (2,−1,3) satisfies 2(x−1)−y+3(z−2)=0.

A vector lying in the plane is perpendicular to the normal, not parallel to it; one point alone does not determine a unique plane.

The vector or cross product a×b is perpendicular to both a and b, with magnitude |a×b|=|a||b|sinθ. Its magnitude is the area of the parallelogram spanned by the two vectors.

The order reverses direction: a×b=−(b×a). It is zero when vectors are parallel, so it can test parallelism and construct a plane normal.

(1,0,0)×(0,2,0)=(0,0,2), whose magnitude 2 is the area of the rectangle formed by the vectors.

The cross product is not a scalar dot product, and changing the order is not harmless.

A line is r=a+λd; a plane is n·(r−b)=0. A line is parallel to a plane when n·d=0, lies in it when a point also satisfies the plane equation, and intersects it otherwise.

To find a line–plane intersection, substitute the line into the plane equation to solve λ. For two planes, the cross product of their normals gives the direction of their line of intersection when the normals are not parallel.

For r=(1,0,0)+λ(1,2,−1) and plane x+y+z=3, substitution gives 1+λ+2λ−λ=3, so λ=2/2=1 and the intersection point is (2,2,−1).

Parallel normals do not prove two planes are the same; compare a point or the constants, and keep line direction distinct from position.