Source: Midterm I Practice Exam
Difficulty: Intermediate | Prerequisites: Familiarity with vectors in 2D, basic trigonometry, and the algebra of 3D coordinates.
Vectors, dot products, and cross products form the backbone of multivariable calculus and 3D geometry. Before you can work with curves, surfaces, or planes in three dimensions, you need fluency with these operations. This material sits right at the start of Math 126 (Calculus III) and everything that follows, from parametric curves to partial derivatives, leans on it. If you are comfortable adding, scaling, and taking magnitudes of vectors in 2D, extending those ideas to 3D is straightforward, and the dot and cross products are the two new tools you pick up here.
The dot product measures how much two vectors point in the same direction and gives you angles. The cross product produces a new vector perpendicular to both inputs and its magnitude equals the area of the parallelogram they span. Together, these two operations let you find angles, areas, projections, and normal vectors in 3D.
Dot product (scalar product)
For vectors a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩, the dot product is a · b = a₁b₁ + a₂b₂ + a₃b₃. Equivalently, a · b = |a||b| cos θ, where θ is the angle between them.
In simple terms, this means: a single number that tells you how aligned two vectors are. If it is zero, they are perpendicular.
Cross product (vector product)
For vectors a and b in 3D, the cross product a × b is the vector computed via the determinant of a 3×3 matrix with i, j, k in the first row. Its magnitude is |a||b| sin θ, and it points perpendicular to both a and b (direction given by the right-hand rule).
Think of it as: a new vector that sticks straight out of the plane formed by the two input vectors. Its length equals the area of their parallelogram.
Midpoint of a segment
The midpoint of the segment from P(x₁, y₁, z₁) to Q(x₂, y₂, z₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2).
In simple terms, this means: average the coordinates.
Parallelogram diagonal bisection
In any parallelogram ABDC, the diagonals AD and BC bisect each other, meaning they cross at their shared midpoint.
Think of it as: the diagonals cut each other exactly in half.
Given three corners A, B, D of parallelogram ABCD (with AB ∥ DC and AD ∥ BC), find C.
The vector from A to B equals the vector from D to C.
So AB = B - A, and C = D + AB.
Example: A(2, -1, 3), B(5, -2, 2), D(3, 2, 1). AB = ⟨3, -1, -1⟩. Then C = (3 + 3, 2 + (-1), 1 + (-1)) = (6, 1, 0).
The diagonals of a parallelogram bisect each other. So the intersection point is the midpoint of either diagonal.
Diagonal from A(2, -1, 3) to C(6, 1, 0): midpoint = ((2+6)/2, (-1+1)/2, (3+0)/2) = (4, 0, 3/2).
Check: diagonal from B(5, -2, 2) to D(3, 2, 1): midpoint = ((5+3)/2, (-2+2)/2, (2+1)/2) = (4, 0, 3/2). Confirmed.
To find the angle θ between vectors u and v:
Compute u · v.
Compute |u| and |v|.
cos θ = (u · v) / (|u| |v|).
θ = arccos(result). If the question asks for the acute angle, take the acute version.
Example (diagonals): AC = C - A = ⟨4, 2, -3⟩, BD = D - B = ⟨-2, 4, -1⟩.
AC · BD = (4)(-2) + (2)(4) + (-3)(-1) = -8 + 8 + 3 = 3.
|AC| = √(16 + 4 + 9) = √29, |BD| = √(4 + 16 + 1) = √21.
cos θ = 3 / (√29 · √21) = 3/√609.
θ = arccos(3/√609). Since this is already acute, that is the answer.
The area of the parallelogram spanned by vectors u and v is |u × v|.
Use the adjacent side vectors, not the diagonals. Here AB = ⟨3, -1, -1⟩ and AD = D - A = ⟨1, 3, -2⟩.
AB × AD via the determinant:
i((-1)(-2) - (-1)(3)) - j((3)(-2) - (-1)(1)) + k((3)(3) - (-1)(1))
= i(2 + 3) - j(-6 + 1) + k(9 + 1)
= ⟨5, 5, 10⟩.
|⟨5, 5, 10⟩| = √(25 + 25 + 100) = √150 = 5√6.
Area = 5√6.
\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\text{Midpoint} = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2},\; \frac{z_1 + z_2}{2}\right)\text{Area of parallelogram} = |\mathbf{u} \times \mathbf{v}|The cross product is how engineers compute torque (force applied at a distance from a pivot). It is also how 3D graphics engines determine which direction a surface faces, which is essential for lighting calculations. The dot product underpins everything from calculating work done by a force to measuring similarity between data vectors in machine learning.
Students often use the diagonal vectors to compute the area of a parallelogram. The area formula uses the adjacent side vectors, not the diagonals. Using the diagonals gives you half the area (the area of a triangle), scaled incorrectly.
Students sometimes forget to take the absolute value or the acute angle when asked for "the acute angle." If your arccos gives an obtuse angle, subtract from π.
The cross product is not commutative. a × b = -(b × a). Swapping the order flips the sign. The magnitude stays the same, but the direction reverses.
The dot product gives a scalar. The cross product gives a vector. Mixing these up in a formula is a common source of errors.
⚠️ Finding a missing vertex of a parallelogram by vector addition is a standard exam question. Know the pattern: C = D + AB (or equivalently C = B + AD).
⚠️ You must give exact answers, not decimal approximations. Leave answers as arccos(3/√609), not 83.0°.
⚠️ The cross product determinant is the most common place for sign errors. Practise the 3×3 determinant expansion until it is automatic.
⚠️ When asked for the area of a parallelogram, the answer is |u × v|, not ½|u × v|. The half factor is for triangles.
True or false: a · b = b · a for all vectors. (True.)
True or false: a × b = b × a for all vectors. (False. It equals -(b × a).)
If a · b = 0, the vectors are ___. (Perpendicular / orthogonal.)
The area of a parallelogram with sides u and v equals ___. (|u × v|.)
True or false: the cross product of two parallel vectors is the zero vector. (True.)
Q: A parallelogram has vertices A(1, 0, 2), B(4, 1, 0), and D(2, 3, 1). Find vertex C.
A: AB = ⟨3, 1, -2⟩. C = D + AB = (2+3, 3+1, 1-2) = (5, 4, -1).
Q: Find the area of the parallelogram with sides u = ⟨1, 2, 0⟩ and v = ⟨0, 1, 3⟩.
A: u × v = ⟨(2)(3) - (0)(1), (0)(0) - (1)(3), (1)(1) - (2)(0)⟩ = ⟨6, -3, 1⟩. Area = |⟨6, -3, 1⟩| = √(36 + 9 + 1) = √46.
Q: Find the angle between ⟨1, 0, 1⟩ and ⟨0, 1, 1⟩.
A: Dot product = 0 + 0 + 1 = 1. Magnitudes = √2 and √2. cos θ = 1/2. θ = π/3 (60°).
Q: If two vectors are parallel, what is their cross product?
A: The zero vector ⟨0, 0, 0⟩, because sin 0 = 0.
The cross product reappears when you find normal vectors to planes (Topic 2 in these notes). The dot product is central to projections, which feed into arc-length and curvature calculations later in the course. Understanding parallelogram area via the cross product extends naturally to computing the area of surfaces in 3D (surface integrals in Math 324).
Tags: dot product, scalar product, cross product, vector product, parallelogram area, midpoint formula, 3D vectors, angle between vectors, determinant method, right-hand rule, vector magnitude, Math 126, Calculus III, UW, multivariable calculus, vector operations