โž—Calculus II

Key Concepts of Vectors in 3D Space

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Why This Matters

Vectors in 3D space form the foundation for everything you'll encounter in multivariable calculus, from describing curves and surfaces to calculating work, flux, and circulation in later chapters. When you're working with parametric curves, line integrals, or surface integrals, you're building directly on the vector operations and geometric representations covered here. The concepts of dot products, cross products, projections, and vector equations show up repeatedly in both computational problems and theoretical questions.

You're being tested on more than just formulas. Exam questions will ask you to interpret what a dot product of zero means geometrically, or to set up a vector equation for a line through two points. Don't just memorize the mechanics: know what concept each operation illustrates and when to apply it. If you understand why the cross product produces a perpendicular vector or how projections decompose forces, you'll handle FRQs with confidence.


Foundational Vector Operations

These operations are the building blocks for everything else. Master them first, and the rest of the topic becomes straightforward.

Definition of Vectors in 3D Space

A vector in 3D is an ordered triplet written as โŸจx,y,zโŸฉ\langle x, y, z \rangle, representing a displacement (direction and magnitude) rather than a single point. Its magnitude (or length) is calculated as:

โˆฃvโˆฃ=x2+y2+z2|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}

This is just the 3D version of the Pythagorean theorem. Geometrically, you can visualize vectors as arrows in space, which makes operations like addition (tip-to-tail) and scalar multiplication (stretching or shrinking the arrow) intuitive.

A unit vector has magnitude 1 and points in a specific direction. To find the unit vector in the direction of v\mathbf{v}, divide by its magnitude: v^=vโˆฃvโˆฃ\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}. This comes up constantly when you need direction without caring about length.

Vector Operations (Addition, Subtraction, Scalar Multiplication)

All three operations work component-wise. For A=โŸจa1,a2,a3โŸฉ\mathbf{A} = \langle a_1, a_2, a_3 \rangle and B=โŸจb1,b2,b3โŸฉ\mathbf{B} = \langle b_1, b_2, b_3 \rangle:

  • Addition: A+B=โŸจa1+b1,a2+b2,a3+b3โŸฉ\mathbf{A} + \mathbf{B} = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle
  • Subtraction: Aโˆ’B=โŸจa1โˆ’b1,a2โˆ’b2,a3โˆ’b3โŸฉ\mathbf{A} - \mathbf{B} = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle
  • Scalar multiplication: kA=โŸจka1,ka2,ka3โŸฉk\mathbf{A} = \langle ka_1, ka_2, ka_3 \rangle, where a negative kk reverses direction

Linear combinations of vectors (like c1A+c2Bc_1\mathbf{A} + c_2\mathbf{B}) form the basis for parametric representations of lines and planes. FRQs often ask you to express a new vector as a linear combination, requiring both operations together.

Compare: Addition vs. Scalar Multiplication: both are component-wise, but addition combines two vectors while scalar multiplication stretches one. Addition changes both magnitude and direction in general; scalar multiplication changes only magnitude (and possibly flips direction).


Products That Reveal Geometry

The dot and cross products aren't just computational tools. They encode geometric information about angles, alignment, and perpendicularity. Understanding their geometric meaning is more important than memorizing formulas.

Dot Product and Its Properties

The dot product takes two vectors and returns a scalar (a number, not a vector).

  • Algebraic formula: Aโ‹…B=a1b1+a2b2+a3b3\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3
  • Geometric formula: Aโ‹…B=โˆฃAโˆฃโˆฃBโˆฃcosโกฮธ\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos\theta

The geometric formula is where the real insight lives. Since the result depends on cosโกฮธ\cos\theta:

  • Positive dot product โ†’ angle between vectors is acute (less than 90ยฐ)
  • Zero dot product โ†’ vectors are perpendicular (orthogonal)
  • Negative dot product โ†’ angle between vectors is obtuse (greater than 90ยฐ)

To find the angle between two vectors, rearrange: cosโกฮธ=Aโ‹…BโˆฃAโˆฃโˆฃBโˆฃ\cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}. Key properties include commutativity (Aโ‹…B=Bโ‹…A\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}) and distributivity over addition.

Cross Product and Its Properties

The cross product takes two vectors and returns a vector perpendicular to both inputs.

  • Magnitude: โˆฃAร—Bโˆฃ=โˆฃAโˆฃโˆฃBโˆฃsinโกฮธ|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin\theta, which equals the area of the parallelogram formed by the two vectors
  • Direction: determined by the right-hand rule (curl fingers from A\mathbf{A} toward B\mathbf{B}; your thumb points in the direction of Aร—B\mathbf{A} \times \mathbf{B})

To compute Aร—B\mathbf{A} \times \mathbf{B}, use the determinant method:

Aร—B=โˆฃijka1a2a3b1b2b3โˆฃ\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}

This expands to โŸจa2b3โˆ’a3b2,โ€…โ€Ša3b1โˆ’a1b3,โ€…โ€Ša1b2โˆ’a2b1โŸฉ\langle a_2b_3 - a_3b_2,\; a_3b_1 - a_1b_3,\; a_1b_2 - a_2b_1 \rangle.

The cross product is anti-commutative: Aร—B=โˆ’Bร—A\mathbf{A} \times \mathbf{B} = -\mathbf{B} \times \mathbf{A}. Order matters. Also note: if Aร—B=0\mathbf{A} \times \mathbf{B} = \mathbf{0}, the vectors are parallel (since sinโกฮธ=0\sin\theta = 0).

Projections of Vectors

The projection of A\mathbf{A} onto B\mathbf{B} gives you the component of A\mathbf{A} that points in the direction of B\mathbf{B}:

projBA=Aโ‹…BโˆฃBโˆฃ2B\text{proj}_{\mathbf{B}}\mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2}\mathbf{B}

This is a vector. The scalar version (the signed length of the projection) is called the scalar component:

compBA=Aโ‹…BโˆฃBโˆฃ\text{comp}_{\mathbf{B}}\mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}

Projection lets you decompose any vector into a parallel part and a perpendicular part relative to another vector. The perpendicular part is just Aโˆ’projBA\mathbf{A} - \text{proj}_{\mathbf{B}}\mathbf{A}. Physical applications include resolving forces along a direction and calculating work: W=Fโ‹…dW = \mathbf{F} \cdot \mathbf{d}.

Compare: Dot Product vs. Cross Product: the dot product measures alignment (returns a scalar, uses cosine), while the cross product measures perpendicularity (returns a vector, uses sine). If an FRQ asks about angles, use dot product. If it asks for a normal vector or area, use cross product.


Equations of Lines and Planes

These representations let you describe geometric objects algebraically. The key distinction: lines need one parameter, planes need two (or equivalently, one normal vector constraint).

Vector Equations of Lines

A line in 3D is determined by a point and a direction. The vector equation is:

r(t)=r0+tv\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}

Here r0\mathbf{r}_0 is the position vector of a known point on the line, v\mathbf{v} is the direction vector, and the parameter tt ranges over all real numbers. Each value of tt gives a different point on the line.

To find the direction vector between two points PP and QQ, just subtract: v=PQโ†’=Qโˆ’P\mathbf{v} = \overrightarrow{PQ} = Q - P.

Parametric Equations of Lines

The parametric form breaks the vector equation into components:

x=x0+at,y=y0+bt,z=z0+ctx = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct

where โŸจa,b,cโŸฉ\langle a, b, c \rangle is the direction vector and (x0,y0,z0)(x_0, y_0, z_0) is a point on the line.

Symmetric equations eliminate tt by setting the expressions equal:

xโˆ’x0a=yโˆ’y0b=zโˆ’z0c\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}

Watch out: if any direction component is zero (say a=0a = 0), you can't divide by it. Instead, that coordinate is constant (x=x0x = x_0), and you write the symmetric equation using only the remaining components.

Converting between vector, parametric, and symmetric forms is a common exam task. Practice going in all directions.

Compare: Vector vs. Parametric Form for Lines: they contain the same information, but vector form is more compact for theoretical work, while parametric form is better for finding specific coordinates or checking if a point lies on the line.

Vector Equations of Planes

A plane is determined by a point and a normal vector (a vector perpendicular to the plane). The point-normal form is:

nโ‹…(rโˆ’r0)=0\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0

This says: any vector lying in the plane is perpendicular to the normal. Expanding this gives the scalar equation:

Ax+By+Cz=DAx + By + Cz = D

where โŸจA,B,CโŸฉ\langle A, B, C \rangle is the normal vector and D=Ax0+By0+Cz0D = Ax_0 + By_0 + Cz_0.

Alternatively, the two-parameter form uses two non-parallel direction vectors u\mathbf{u} and w\mathbf{w} that lie in the plane:

r(s,t)=r0+su+tw\mathbf{r}(s, t) = \mathbf{r}_0 + s\mathbf{u} + t\mathbf{w}

Parametric Equations of Planes

The two-parameter vector equation above expands into three parametric equations (one each for xx, yy, zz) as functions of ss and tt.

The direction vectors u\mathbf{u} and w\mathbf{w} span the plane. Their cross product gives the normal vector: n=uร—w\mathbf{n} = \mathbf{u} \times \mathbf{w}. This is how you convert from parametric form to scalar form:

  1. Compute n=uร—w\mathbf{n} = \mathbf{u} \times \mathbf{w} to get โŸจA,B,CโŸฉ\langle A, B, C \rangle
  2. Pick any known point on the plane (use r0\mathbf{r}_0)
  3. Substitute into Ax+By+Cz=DAx + By + Cz = D to find DD

Compare: Lines vs. Planes: lines use one parameter and one direction vector; planes use two parameters and two direction vectors (or equivalently, one normal vector constraint). This distinction drives how you set up intersection problems.


Distance Calculations

Distance formulas combine vector operations with geometric insight. Each formula has a specific structure, so know which one applies to each situation.

Distance Between Points, Lines, and Planes

Point-to-point distance is the direct 3D extension of the Pythagorean theorem:

d=(x2โˆ’x1)2+(y2โˆ’y1)2+(z2โˆ’z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Point-to-plane distance uses the scalar equation of the plane. If the plane is Ax+By+Cz+D=0Ax + By + Cz + D = 0 and the point is (x0,y0,z0)(x_0, y_0, z_0):

d=โˆฃAx0+By0+Cz0+DโˆฃA2+B2+C2d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}

Note the plane equation must be in the form Ax+By+Cz+D=0Ax + By + Cz + D = 0 (with DD on the left side) for this formula to work directly.

Point-to-line distance requires a cross product. If the line passes through point PP with direction v\mathbf{v}, and you want the distance from point QQ to the line:

d=โˆฃPQโ†’ร—vโˆฃโˆฃvโˆฃd = \frac{|\overrightarrow{PQ} \times \mathbf{v}|}{|\mathbf{v}|}

The cross product gives the area of the parallelogram formed by PQโ†’\overrightarrow{PQ} and v\mathbf{v}, and dividing by the base length โˆฃvโˆฃ|\mathbf{v}| gives the height, which is the perpendicular distance.

Compare: Point-to-Plane vs. Point-to-Line Distance: both measure perpendicular distance, but the plane formula is a direct substitution while the line formula requires a cross product. On FRQs, identify which object you're measuring distance to before selecting your approach.


Vector-Valued Functions and Motion

This section bridges vectors with calculus. Vector-valued functions let you describe paths through space and analyze motion using derivatives.

Vector-Valued Functions and Space Curves

A vector-valued function assigns a position vector to each value of a parameter tt:

r(t)=โŸจx(t),y(t),z(t)โŸฉ\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle

The path traced out by r(t)\mathbf{r}(t) as tt varies is called a space curve. Common examples include helices (r(t)=โŸจcosโกt,sinโกt,tโŸฉ\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle), circles in 3D, and curves formed by the intersection of two surfaces.

Differentiation connects vector geometry to motion:

  • rโ€ฒ(t)\mathbf{r}'(t) gives the velocity vector, which is tangent to the curve at each point
  • rโ€ฒโ€ฒ(t)\mathbf{r}''(t) gives the acceleration vector
  • Differentiation is done component-wise: rโ€ฒ(t)=โŸจxโ€ฒ(t),yโ€ฒ(t),zโ€ฒ(t)โŸฉ\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle

A parametric line (r(t)=r0+tv\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}) is the simplest vector-valued function since its components are linear in tt. Curves with nonlinear components require more careful analysis of tangent vectors and arc length.

Compare: Parametric Lines vs. General Vector-Valued Functions: a line is the simplest vector-valued function (linear in tt), while curves like r(t)=โŸจcosโกt,sinโกt,tโŸฉ\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle involve nonlinear components. Differentiation techniques apply to both, but curves require more careful analysis of tangent vectors.


Quick Reference Table

ConceptBest Examples
Scalar result from vectorsDot product, magnitude calculation
Vector result from vectorsCross product, projection, addition
Perpendicularity testDot product equals zero
Parallelism testCross product equals zero vector
Normal vector to a planeCross product of two vectors in the plane
Line representationVector equation r(t)=r0+tv\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}, parametric equations
Plane representationPoint-normal form, scalar equation Ax+By+Cz=DAx + By + Cz = D
Distance calculationsPoint-to-point, point-to-plane, point-to-line formulas
Motion in 3DVector-valued functions, derivatives for velocity/acceleration

Self-Check Questions

  1. What geometric condition does Aโ‹…B=0\mathbf{A} \cdot \mathbf{B} = 0 indicate, and how does this differ from what Aร—B=0\mathbf{A} \times \mathbf{B} = \mathbf{0} tells you?

  2. Given two points in 3D space, describe the steps to write both the vector equation and parametric equations for the line through them.

  3. Compare and contrast finding the distance from a point to a plane versus the distance from a point to a line. What vector operations does each require?

  4. If you need a vector perpendicular to a plane, which product do you use and what inputs do you need? How does this connect to the scalar equation of a plane?

  5. Explain how the projection projBA\text{proj}_{\mathbf{B}}\mathbf{A} relates to the dot product, and give one physical situation where this decomposition is useful.