Multivariable Calculus

Vectors and Dot Products

01Vectors, Components, and Length in Space 02Scaling and Adding Vectors: The Parallelogram Rule 03The Dot Product and the Law of Cosines 04Finding Angles in Space With the Dot Product 05The Perpendicularity Test and the Normal Vector to a Plane Problem set0/10 Problem set 20/10 MIT problem set0/2 Practice∞

Determinants and the Cross Product

01The Component of a Vector Along Any Unit Direction 02The 2x2 Determinant and the Area of a Parallelogram 03The 3x3 Determinant and the Volume of a Box 04The Cross Product and the Right-Hand Rule 05The Triple Product and the Volume of a Parallelepiped Problem set0/10 Problem set 20/10 MIT problem set0/2 Practice∞

Matrices and Inverse Matrices

01Why Swapping a Cross Product Flips Its Sign 02The Equation of a Plane Through Three Points 03Matrix Multiplication: Rows Dotted With Columns 04Matrices as Transformations: Identity and Rotation 05The Inverse Matrix and Solving AX = B 06Computing a 3x3 Inverse: Minors, Cofactors, Adjoint Problem set0/10 Problem set 20/10 MIT problem set0/2 Practice∞

Planes and Square Systems

01Planes From a Point and a Normal Vector 02Reading the Normal Vector Off a Plane Equation 03Three Planes, One Point: Solving a 3x3 System 04Invertibility, det A, and the Homogeneous System 05When det A = 0: Coplanar Normals and a Line of Solutions Problem set0/10 Problem set 20/10 Practice∞

The Perpendicularity Test and the Normal Vector to a Plane

Why does the single equation x + 2y + 3z = 0 carve out a plane, and where in it is the normal vector hiding?

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Your summary note

1. The zero dot product orthogonality test
  State that two vectors are perpendicular exactly when their dot product is zero, and derive it from $|\vec{A}||\vec{B}|\cos\theta = 0$ giving $\cos\theta = 0$ and $\theta = \pi/2$ .
2. Reading $x + 2y + 3z = 0$ as a dot product
  Write $\vec{A} = \langle 1,2,3 \rangle$ and $\overrightarrow{OP} = \langle x,y,z \rangle$ , and show the left side equals $\vec{A}\cdot\overrightarrow{OP}$ so the equation becomes $\vec{A}\cdot\overrightarrow{OP} = 0$ .
3. Solution set is the plane with normal vector $\langle 1,2,3 \rangle$
  Conclude the solutions are the points whose position vector is perpendicular to $\vec{A}$ , forming the plane through the origin with normal $\vec{A}$ , mirroring how vectors perpendicular to a vertical vector fill the horizontal plane.
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