Ludium
Sign In
Multivariable Calculus
Vectors and Dot Products
01Vectors, Components, and Length in Space02Scaling and Adding Vectors: The Parallelogram Rule03The Dot Product and the Law of Cosines04Finding Angles in Space With the Dot Product05The Perpendicularity Test and the Normal Vector to a PlaneProblem set0/10Problem set 20/10MIT problem set0/2Practice∞
Determinants and the Cross Product
01The Component of a Vector Along Any Unit Direction02The 2x2 Determinant and the Area of a Parallelogram03The 3x3 Determinant and the Volume of a Box04The Cross Product and the Right-Hand Rule05The Triple Product and the Volume of a ParallelepipedProblem set0/10Problem set 20/10MIT problem set0/2Practice∞
Matrices and Inverse Matrices
01Why Swapping a Cross Product Flips Its Sign02The Equation of a Plane Through Three Points03Matrix Multiplication: Rows Dotted With Columns04Matrices as Transformations: Identity and Rotation05The Inverse Matrix and Solving AX = B06Computing a 3x3 Inverse: Minors, Cofactors, AdjointProblem set0/10Problem set 20/10MIT problem set0/2Practice∞
Planes and Square Systems
01Planes From a Point and a Normal Vector02Reading the Normal Vector Off a Plane Equation03Three Planes, One Point: Solving a 3x3 System04Invertibility, det A, and the Homogeneous System05When det A = 0: Coplanar Normals and a Line of SolutionsProblem set0/10Problem set 20/10Practice∞

Planes From a Point and a Normal Vector

Why does demanding a single right angle between a normal vector and a displacement give you the whole equation ax + by + cz = d?


Loading…

←Previous Computing a 3x3 Inverse: Minors, Cofactors, AdjointNext Reading the Normal Vector Off a Plane Equation →

Your summary note

    1. 1

      Two parallel planes from one normal vector N⃗\vec{N}N

      Work both examples with N⃗=⟨1,5,10⟩\vec{N} = \langle 1,5,10\rangleN=⟨1,5,10⟩: the plane through the origin from OP→⋅N⃗=0\overrightarrow{OP} \cdot \vec{N} = 0OP⋅N=0, and the plane through P0=(2,1,−1)P_0 = (2,1,-1)P0​=(2,1,−1) from P0P→⋅N⃗=0\overrightarrow{P_0P} \cdot \vec{N} = 0P0​P​⋅N=0.

    2. 2

      The coefficients aaa, bbb, ccc as a normal vector

      Compare x+5y+10z=0x + 5y + 10z = 0x+5y+10z=0 with x+5y+10z=−3x + 5y + 10z = -3x+5y+10z=−3, record that aaa, bbb, ccc give N⃗\vec{N}N, and that only ddd changes between parallel planes.

    3. 3

      Getting ddd by evaluating at a known point

      Record the shortcut for the constant, and check it on the example by substituting P0=(2,1,−1)P_0 = (2,1,-1)P0​=(2,1,−1) into x+5y+10zx + 5y + 10zx+5y+10z to recover −3-3−3.

    Attempt 1 of 2