Difficulty: Intermediate | Prerequisites: Calculus I and II (derivatives, integrals, parametric curves), basic linear algebra (vectors, dot product)
This material covers the first major block of Calculus III: working in three dimensions. You are moving from single-variable calculus into multivariable territory, and before you can differentiate or integrate functions of several variables, you need the geometric toolkit: vectors, planes, curves in space, and the surfaces they live on. Everything here feeds directly into partial derivatives, gradients, and multiple integrals later in the course.
TL;DR: Vectors in 3D let you define planes (via cross products and normal vectors), measure angles between geometric objects, compute arc lengths of space curves, and classify surfaces. The dot product measures alignment; the cross product gives you a perpendicular direction and an area. Linearisation (covered in Doc 2) depends on understanding these objects geometrically.
Normal vector
A vector perpendicular to a plane (or surface). Every plane in 3D has a normal direction, and the equation of the plane is built from it. In simple terms: the arrow that sticks straight out of a flat surface.
Cross product (a x b)
An operation on two 3D vectors that returns a third vector perpendicular to both, with magnitude equal to the area of the parallelogram they span. Think of it as: the tool that finds "the direction sticking out" from two directions lying in a surface.
Dot product (a . b)
An operation on two vectors that returns a scalar: a . b = |a||b| cos(theta). Zero when the vectors are perpendicular. In simple terms: it measures how much two vectors point in the same direction.
Direction cosines
The cosines of the angles a vector makes with the x-, y-, and z-axes. For a vector v = (v1, v2, v3), these are v1/|v|, v2/|v|, v3/|v|. They always satisfy cos^2(theta_1) + cos^2(theta_2) + cos^2(theta_3) = 1. Think of it as: the vector's unit-vector components, read as cosines.
Arc length
The total distance travelled along a curve r(t) from t = a to t = b, computed as the integral of |r'(t)| dt. In simple terms: if you straightened the curve into a line, how long would it be.
Linearisation (tangent plane approximation)
The best linear approximation to a function near a point. For f(x, y) near (a, b): L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b). Think of it as: zooming in until the surface looks flat, then using that flat plane to estimate nearby values.
Parabolic cylinder
A surface that is a parabola in one cross-section and extends infinitely (unchanged) in the perpendicular direction. In simple terms: take a parabola and slide it sideways forever, like a curved roof.
Level set (contour)
The set of all points where a function takes a single constant value, e.g. f(x, y) = c. Think of it as: a contour line on a topographic map, connecting all points at the same elevation.
The standard equation of a plane is n . (r - r_0) = 0, where n is the normal vector and r_0 is any known point on the plane. The problem is finding n.
Step 1: Take any two pairs from the three given points to form two vectors that lie in the plane. For example, given points P, Q, R: form v1 = Q - P and v2 = R - P.
Step 2: Compute n = v1 x v2. This cross product is perpendicular to both vectors and therefore perpendicular to the plane.
Step 3: Plug n and any one of the three points into the plane equation n . (r - r_0) = 0, then expand and simplify.
Worked example (Midterm Q1a): Points (0, 3, 0), (-3, 0, 0), (2, 6, 1).
Form v1 = (3, 3, 0) and v2 = (2, 3, 1). Cross product:
\mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 3 & 0 \\ 2 & 3 & 1 \end{vmatrix} = \langle 3, -3, 3 \rangleSimplify to n = (1, -1, 1). Using the point (-3, 0, 0):
(1)(x + 3) + (-1)(y) + (1)(z) = 0, giving x - y + z = -3.
The angle between two planes is the angle between their normal vectors. If n1 and n2 are the normals:
\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1|\,|\mathbf{n}_2|}If this gives a negative value, the computed angle is the supplement (pi minus the acute angle between the planes). Both the angle and its supplement are acceptable answers, because the normal can point in either direction.
Finding the normal of a plane defined by a point and a line: Extract the direction vector of the line (that vector lies in the plane). Then find a second vector in the plane by connecting two known points on the plane. Cross the two to get the normal.
Worked example (Midterm Q1b): Plane P2 contains the point (0, 2, 1) and the line l(t) = (2t, t, 1 + 3t). The direction vector of the line is (2, 1, 3). The line passes through (0, 0, 1) at t = 0, so a second vector in P2 is (0, 2, 1) - (0, 0, 1) = (0, 1, 0). Then:
n2 = (2, 1, 3) x (0, 1, 0) = (-3, 0, 2).
With n1 = (1, -1, 1) from the earlier problem:
cos(theta) = -1 / sqrt(3 * 13) = -1 / sqrt(39), so theta = arccos(-1/sqrt(39)).
The distance from a point (x0, y0, z0) to a plane ax + by + cz = d is:
D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}The shortest distance from a plane to a sphere equals the distance from the plane to the sphere's centre, minus the radius.
Exam example (Q2a): Distance from x + y + z = 0 to the sphere of radius 1 centred at (1, 2, 3).
Distance from centre to plane = |1 + 2 + 3| / sqrt(3) = 6/sqrt(3) = 2sqrt(3).
Shortest distance to the sphere = 2sqrt(3) - 1. The statement "the distance equals 2sqrt(3) - 1" is TRUE.
A vector w is orthogonal to both u and v if and only if w is parallel to u x v. This means w = k(u x v) for some scalar k.
To check whether some vector (5, a, b) can be orthogonal to both (1, 2, 3) and (1, -2, 1), compute the cross product:
(1, 2, 3) x (1, -2, 1) = (2 + 6, 3 - 1, -2 - 2) = (8, 2, -4).
Any vector orthogonal to both must be a scalar multiple of (8, 2, -4), i.e. of the form (8k, 2k, -4k). Setting the first component to 5 gives k = 5/8, so a = 10/8 = 5/4 and b = -20/8 = -5/2. Values of a and b do exist. The statement "there are no values" is FALSE.
For any two vectors u and v, the parallelogram law states:
2|\mathbf{u}|^2 + 2|\mathbf{v}|^2 = |\mathbf{u} - \mathbf{v}|^2 + |\mathbf{u} + \mathbf{v}|^2This is an identity, meaning it holds for all vectors with no exceptions. It follows from expanding both sides using the dot product.
The claim that there exist u, v with 2|v|^2 + 2|u|^2 > |u - v|^2 + |u + v|^2 is FALSE, because the two sides are always exactly equal.
Given a plane ax + by + cz = d, substitute a point into the left side. Two points lie on the same side if and only if the substitutions give the same sign (both positive or both negative). If one is positive and the other negative, they are on opposite sides.
Exam example (Q2f): Plane x - y + z = 2. Point A = (1, 2, 2): 1 - 2 + 2 = 1 > 0. Point B = (2, 1, 2): 2 - 1 + 2 = 3 > 0. Both positive, so same side? The statement says FALSE. Re-checking: the plane is x - y + z = 2. For A: 1 - 2 + 2 = 1, so A gives value 1 (positive side relative to the constant 2, but 1 < 2). For B: 2 - 1 + 2 = 3 > 2.
The proper test is the sign of (ax + by + cz - d). For A: 1 - 2 = -1 < 0. For B: 3 - 2 = 1 > 0. Opposite signs, so the points are on opposite sides. The statement is FALSE.
For a vector v = (v1, v2, v3), the direction cosines are the cosines of the angles theta_1, theta_2, theta_3 that v makes with the x-, y-, and z-axes respectively:
\cos\theta_1 = \frac{v_1}{|\mathbf{v}|}, \quad \cos\theta_2 = \frac{v_2}{|\mathbf{v}|}, \quad \cos\theta_3 = \frac{v_3}{|\mathbf{v}|}These are simply the components of the unit vector v/|v|.
Key identity: cos^2(theta_1) + cos^2(theta_2) + cos^2(theta_3) = 1.
This follows directly from the fact that (v1^2 + v2^2 + v3^2) / |v|^2 = |v|^2 / |v|^2 = 1. It holds for any nonzero vector, regardless of its direction.
For a curve r(t) = (x(t), y(t), z(t)) from t = a to t = b:
s = \int_a^b |\mathbf{r}'(t)|\, dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\, dtThe strategy is:
Compute r'(t) component by component.
Find |r'(t)| and simplify. Look for perfect squares under the radical; many exam problems are designed so the expression under the square root factors cleanly.
Integrate.
Worked example (Midterm Q4a): r(t) = (t + 1, (2sqrt(2)/3) t^(3/2) + 8, t^2/2) for 1 <= t <= 3.
r'(t) = (1, sqrt(2t), t).
|r'(t)| = sqrt(1 + 2t + t^2) = sqrt((1 + t)^2) = |1 + t| = 1 + t (since t >= 1 > 0).
Arc length = integral from 1 to 3 of (1 + t) dt = [t + t^2/2] from 1 to 3 = (3 + 9/2) - (1 + 1/2) = 6.
Some exam problems define a surface implicitly by a geometric condition (e.g. "all points equidistant from a point and a line") and ask you to find its equation and classify it.
The general approach:
Write the distance from (x, y, z) to the given point using the standard distance formula.
Write the distance from (x, y, z) to the given line. For a line in the xy-plane described by ax + by = 0, z = 0: first find the perpendicular distance in the xy-plane using |ax + by| / sqrt(a^2 + b^2), then account for the z-component using the Pythagorean theorem to get sqrt(l^2 + z^2).
Set the two distances equal and simplify algebraically (squaring both sides usually helps).
Worked example (Midterm Q5): Equidistance from (0, 0, 4) and the line 2x + y = 0, z = 0.
d1 = sqrt(x^2 + y^2 + (z - 4)^2).
Distance to the line: l = |2x + y| / sqrt(5), then d2 = sqrt(l^2 + z^2) = sqrt((2x + y)^2/5 + z^2).
Setting d1^2 = d2^2 and simplifying gives z = 2 + (x - 2y)^2 / 40.
This is a parabolic cylinder. Its line of symmetry runs along x = 2y, z = 2, and the parabolic cross-sections open upward in the direction perpendicular to that line.
If |c'(t)| is constant, the curve is traversed at constant speed, but this tells you nothing about the shape of the path. The curve could be a straight line, a circle, a helix, or any other shape in 3D.
The claim that constant |c'(t)| forces the curve to lie in a plane is FALSE. A standard counterexample is the helix r(t) = (cos t, sin t, t), which has |r'(t)| = sqrt(sin^2 t + cos^2 t + 1) = sqrt(2) (constant), but the curve spirals through all of 3D space.
The angle between the position vector r(t) and the tangent vector r'(t) is found from:
cos(alpha) = r(t) . r'(t) / (|r(t)| |r'(t)|).
For the logarithmic spiral r(t) = (e^t cos t, e^t sin t), one can show:
r(t) . r'(t) = e^(2t)(cos t (cos t - sin t) + sin t (sin t + cos t)) = e^(2t)(cos^2 t + sin^2 t) = e^(2t).
|r(t)| = e^t.
|r'(t)| = e^t sqrt((cos t - sin t)^2 + (sin t + cos t)^2) = e^t sqrt(2).
So cos(alpha) = e^(2t) / (e^t * e^t sqrt(2)) = 1/sqrt(2), giving alpha = pi/4, a constant.
The statement that the angle is constant is TRUE. This is a well-known property of logarithmic spirals.
Students often think the angle between two planes can only be found one way. In fact, flipping a normal vector gives the supplementary angle, and both answers are valid.
Students confuse "same side of a plane" tests. You must evaluate the sign of (ax + by + cz - d), not just (ax + by + cz). Forgetting to subtract d changes the answer.
Constant speed does not mean a curve is simple or planar. A helix has constant speed but lives in 3D.
The parallelogram law is an identity, not an inequality. Students sometimes try to find counterexamples, but none exist: 2|u|^2 + 2|v|^2 = |u - v|^2 + |u + v|^2 always.
⚠️ "Find the equation of the plane" is a near-certain exam question. Know the three-point method cold: two vectors, cross product, plug in.
⚠️ True/false questions test conceptual understanding of identities (parallelogram law) and counterexamples (constant speed does not imply planar). These are worth quick points if you know the theory.
⚠️ Arc length problems are designed to simplify cleanly. If your integral looks ugly, re-check whether the expression under the radical factors as a perfect square.
⚠️ Surface classification (ellipsoid, hyperboloid, paraboloid, cylinder) appears frequently. Recognise the standard forms.
⚠️ The distance-from-point-to-plane formula appears in both computation and true/false contexts. Memorise it.
True or false: the cross product of two vectors always has magnitude equal to the area of the triangle they span. (Answer: False, it equals the area of the parallelogram; the triangle is half that.)
Fill in the blank: the sum of the squares of the direction cosines of any nonzero vector equals ___. (Answer: 1.)
True or false: if two points give the same sign when substituted into ax + by + cz, they are on the same side of the plane ax + by + cz = d. (Answer: False, you must evaluate ax + by + cz - d.)
Fill in the blank: the arc length of r(t) from t = a to t = b is the integral of ___. (Answer: |r'(t)| dt.)
True or false: a parabolic cylinder is bounded in all directions. (Answer: False, it extends infinitely along its axis of symmetry.)
Q: Find the equation of the plane through (1, 0, 2), (3, 1, 1), and (0, -1, 4).
A: Form two vectors, e.g. (2, 1, -1) and (-1, -1, 2). Cross product = (2 - 1, -1 - (-4), -2 - (-1)) = (1, 3, -1). Plane: (x - 1) + 3(y) - (z - 2) = 0, i.e. x + 3y - z = -1.
Q: What is the shortest distance from the plane 2x - y + 2z = 6 to the point (1, 1, 1)?
A: D = |2(1) - 1 + 2(1) - 6| / sqrt(4 + 1 + 4) = |2 - 1 + 2 - 6| / 3 = |-3| / 3 = 1.
Q: For the curve r(t) = (3 cos t, 3 sin t, 4t), find the arc length from t = 0 to t = 2pi.
A: r'(t) = (-3 sin t, 3 cos t, 4), |r'(t)| = sqrt(9 + 16) = 5. Arc length = 5 * 2pi = 10pi.
Q: Classify the surface z = x^2 + 4.
A: This is a parabolic cylinder: a parabola in the xz-plane extended infinitely along the y-axis.
Q: Do the points (0, 0, 0) and (2, 3, 1) lie on the same side of the plane x + y + z = 3?
A: For (0, 0, 0): 0 + 0 + 0 - 3 = -3 < 0. For (2, 3, 1): 2 + 3 + 1 - 3 = 3 > 0. Opposite signs, so opposite sides.
Normal vectors reappear when you study gradients: the gradient of a function at a point is the normal to its level surface there. Understanding planes and normals now will make the gradient intuitive later.
Arc length connects directly to line integrals in the second half of the course. The formula is identical, with an extra function inside the integral.
Surface classification (cylinders, quadrics) matters for setting up double and triple integrals, where you need to describe regions bounded by these surfaces.
normal vector, cross product, dot product, plane equation, angle between planes, distance from point to plane, direction cosines, arc length, parametric curve, parabolic cylinder, quadric surface, parallelogram law, orthogonality, logarithmic spiral, constant speed curve, level surface, Calculus III, MATH 215, multivariable calculus, vectors in 3D, space curves