Learning Objectives
Choose vector tools based on the geometric quantity requested.
Combine displacement, cross products, and plane equations in one problem.
Check vector calculations using lengths, dot products, and substitution.
Prerequisites: the three vector-geometry lessons in this section. The aim is to connect the methods before introducing curves and surfaces that change with one or more parameters.
1. Match the Question to the Tool
Question | Tool | Output |
|---|---|---|
How far apart? | Magnitude of displacement | Nonnegative number |
What angle? | Dot product and magnitudes | Angle |
Which direction is perpendicular? | Cross product of independent directions | Vector |
How much area? | Magnitude of cross product | Nonnegative number |
Which points lie in a plane? | Normal dotted with displacement equals zero | Equation |
Checking the output type prevents many errors. An area cannot be a vector, and a normal direction cannot be a scalar. A normal vector need not have length 1 unless the question specifically requests a unit normal.
2. A Complete Geometry Problem
Worked Example: Triangle and Plane
Let A = (1, 0, 0), B = (1, 2, 0), and C = (1, 0, 3). Find the triangle's area and the plane through its vertices.
The edge vectors are AB = ⟨0, 2, 0⟩ and AC = ⟨0, 0, 3⟩. Their cross product is ⟨6, 0, 0⟩.
The cross product has magnitude 6, so the triangle area is 3 square units. Using the cross product as a normal gives 6(x − 1) = 0, or x = 1.
Check the geometry: all three points have x-coordinate 1. Their two perpendicular edges have lengths 2 and 3, so the elementary triangle formula also gives (1/2)(2)(3) = 3.
3. Recognize Degenerate Cases
A zero vector has no direction to normalize. Two parallel nonzero vectors span a line rather than a nonzero-area parallelogram. Three collinear points do not determine a unique plane. These are geometric limitations, not arithmetic failures.
Common Mistakes
A zero cross product means the two vectors are linearly dependent, not perpendicular.
Reversing cross product order changes the normal's direction but not the area or the plane it defines.
Different parameterizations can describe the same line. Check their points and parallel direction vectors before deciding they differ.
4. Mixed Practice
1. Find the distance between (1, 2, 3) and (−1, 2, 6).
2. Find a vector of length 4 in the direction ⟨1, 2, 2⟩.
3. Find the angle between ⟨1, 0, 0⟩ and ⟨1, 1, 0⟩.
4. Find the triangle area for edge vectors ⟨1, 0, 1⟩ and ⟨0, 2, 0⟩.
5. Find the plane through (0, 0, 2) perpendicular to ⟨1, −2, 3⟩. Find the origin's distance to it.
6. Determine whether r(t) = ⟨1 + t, 2 − t, 3⟩ intersects, lies in, or is disjoint from the plane x + y + z = 6.
Worked Solutions
1. The displacement is ⟨−2, 0, 3⟩, so the distance is √13.
2. The given direction has magnitude 3. Multiply it by 4/3 to obtain ⟨4/3, 8/3, 8/3⟩; its squared magnitude is 16.
3. The dot product is 1 and the magnitudes are 1 and √2. Thus cos θ = 1/√2 and θ = π/4 radians, or 45°.
4. The cross product is ⟨−2, 0, 2⟩ with magnitude 2√2. The triangle area is half of that, √2.
5. The equation is x − 2y + 3(z − 2) = 0, or x − 2y + 3z − 6 = 0. The distance from the origin is 6/√14.
6. Substitution gives (1 + t) + (2 − t) + 3 = 6 for every t. Therefore the whole line lies in the plane, rather than meeting it at a single point.
5. Summary
- Use magnitude for length, dot product for angle, and cross product for normals and areas.
- A plane equation tests whether displacements are perpendicular to a chosen normal.
- Check geometric meaning as well as arithmetic, especially in zero-vector and parallel cases.