PAU Mathematics II (Galicia) study materials
PAU Mathematics II (Galicia) study resources

PAU Mathematics II (Galicia): Vectors and analytic geometry in three-dimensional space

PAU Mathematics II (Galicia) · Work through your exam, one topic at a time.

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Vectors and analytic geometry in three-dimensional space

Vectors and analytic geometry in three-dimensional space

One full question per option is usually geometric: relative positions of lines and planes, a perpendicular line or plane, and a distance or an angle. The algebra is short, so marks are lost when the geometry is not justified.

Core rules

  • Products: u.v = u1v1 + u2v2 + u3v3 = |u||v|cos(t); u x v is perpendicular to both vectors and |u x v| = |u||v|sin(t) is the area of the parallelogram they span. The scalar triple product [u,v,w] = u.(v x w) is the signed volume of the parallelepiped, and it vanishes exactly when the three vectors are coplanar.
  • Plane: ax + by + cz = d has normal n = (a,b,c). A plane is fixed by a point and a normal, or by a point and two independent directions, whose cross product is the normal.
  • Line: point plus t times a direction; in implicit form it is the intersection of two planes, and its direction is the cross product of their normals.
  • Relative position of two lines with directions u, v and a vector PQ joining a point of each: rank(u,v,PQ) = 3 means skew lines; rank = 2 means they intersect; rank = 1 means parallel or coincident, decided by testing whether P lies on the second line.
  • Distances: point-plane gives |ax0 + by0 + cz0 + d|/sqrt(a^2 + b^2 + c^2); point-line gives |AP x u|/|u|; skew lines give |[u,v,AP]|/|u x v|. For the angle between a line and a plane use sin(t) = |u.n|/(|u||n|).

Worked example

Find the plane containing the line r: (x,y,z) = (1,0,2) + t(1,1,0) and the point P(0,1,1).

Take the direction u = (1,1,0) and the vector from the point A(1,0,2) of the line to P, AP = (-1,1,-1). The normal is u x AP = (-1,1,2). Hence -1(x - 1) + 1(y - 0) + 2(z - 2) = 0, that is x - y - 2z + 3 = 0. Checking A gives 1 - 0 - 4 + 3 = 0 and checking P gives 0 - 1 - 2 + 3 = 0, so both the line and the point lie in the plane.

Traps that cost marks

Reporting a distance without the absolute value; confusing the normal of a plane with the direction of a line; declaring two lines parallel from their direction vectors without testing a point; using |u x v| where |u.v| belongs (or the reverse) for angles; and leaving the plane equation unverified with the data given.