Finding Angle Between Two Vectors

u→⋅v→=∣∣u→∣∣∗∣∣v→∣∣cosθcosθ=u→⋅v→∣∣u→∣∣∗∣∣v→∣∣\overrightarrow{u} \cdot \overrightarrow{v} = || \overrightarrow{u} || * || \overrightarrow{v} || cos \theta \\ cos \theta = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{|| \overrightarrow{u} || * || \overrightarrow{v} ||}

Direction Cosines

The angle between a vector and the positive x, y, and z axes are alpha, beta, and gamma, respectively.

cosα=v1∣∣v→∣∣cosβ=v2∣∣v→∣∣cosγ=v3∣∣v→∣∣Note that cos2α+cos2β+cos3γ=1cos \alpha = \frac{v_1}{|| \overrightarrow{v} ||} \\ cos \beta = \frac{v_2}{|| \overrightarrow{v} ||} \\ cos \gamma = \frac{v_3}{|| \overrightarrow{v} ||} \\ \text{Note that } cos^2 \alpha + cos^2 \beta + cos^3 \gamma = 1

Projection

Projv→u→=v→u→∣∣v→∣∣2v→=v→u→∣∣v→∣∣v^The vector component orthogonal to u is equal to u→−Projv→u→∣∣Projv→u→∣∣=∣v→⋅u→∣∣∣v→∣∣Proj_{\overrightarrow{v}} \overrightarrow{u} = \frac{\overrightarrow{v} \overrightarrow{u}}{|| \overrightarrow{v} ||^2} \overrightarrow{v} = \frac{\overrightarrow{v} \overrightarrow{u}}{|| \overrightarrow{v} ||} \hat{v} \\ \text{The vector component orthogonal to u is equal to } \overrightarrow{u} - Proj_{\overrightarrow{v}} \overrightarrow{u} \\ || Proj_{\overrightarrow{v}} \overrightarrow{u} || = \frac{| \overrightarrow{v} \cdot \overrightarrow{u} |}{|| \overrightarrow{v} ||}

Work

If a force is being enacted on an object from point P to point Q, then the work can be calculated as follows:

Work=F→⋅PQ→Work = \overrightarrow{F} \cdot \overrightarrow{PQ}

Cross Product

Let u→,v→∈R3. The cross product of u and v is given by the following:u→×v→=(u2v3−u3v2u3v1−u1v3u1v2−u2v1)u→×v→⊥u→,u→×v→⊥v→u→×v→ is orthogonal to the plane containing u and v.u→×v→=det((ijku1u2u3v1v2v3))=i∗det(u2u3v2v3)−j∗det(u1u3v1v3)+k∗det(u1u2v1v2)Note: u→×v→=−(v→×u→) (by the right hand rule)\text{Let } \overrightarrow{u}, \overrightarrow{v} \in \mathbb{R}^3 \text{. The cross product of u and v is given by the following:} \\ \overrightarrow{u} \times \overrightarrow{v} = \begin{pmatrix} u_2 v_3 - u_3 v_2 \\ u_3 v_1 - u_1 v_3 \\ u_1 v_2 - u_2 v_1 \end{pmatrix} \\ \overrightarrow{u} \times \overrightarrow{v} \perp \overrightarrow{u}, \overrightarrow{u} \times \overrightarrow{v} \perp \overrightarrow{v} \\ \overrightarrow{u} \times \overrightarrow{v} \text{ is orthogonal to the plane containing u and v.} \\ \overrightarrow{u} \times \overrightarrow{v} = det(\begin{pmatrix} i & j & k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix}) = i * det\begin{pmatrix} u_2 & u_3 \\ v_2 & v_3 \end{pmatrix} - j * det\begin{pmatrix} u_1 & u_3 \\ v_1 & v_3 \end{pmatrix} + k * det\begin{pmatrix} u_1 & u_2 \\ v_1 & v_2 \end{pmatrix} \\ \text{Note: } \overrightarrow{u} \times \overrightarrow{v} = -(\overrightarrow{v} \times \overrightarrow{u}) \text{ (by the right hand rule)}

Properties of the Cross Product

u→×v→=−(v→×u→)u→×(v→+w→)=u→×v→+u→×w→c(u→×v→)=(cu→)×v→=u→×(cv→)u→×0→=0→×u→=0→u→×u→=0→\overrightarrow{u} \times \overrightarrow{v} = -(\overrightarrow{v} \times \overrightarrow{u}) \\ \overrightarrow{u} \times (\overrightarrow{v} + \overrightarrow{w}) = \overrightarrow{u} \times \overrightarrow{v} + \overrightarrow{u} \times \overrightarrow{w} \\ c (\overrightarrow{u} \times \overrightarrow{v}) = (c \overrightarrow{u}) \times \overrightarrow{v} = \overrightarrow{u} \times (c \overrightarrow{v}) \\ \overrightarrow{u} \times \overrightarrow{0} = \overrightarrow{0} \times \overrightarrow{u} = \overrightarrow{0} \\ \overrightarrow{u} \times \overrightarrow{u} = \overrightarrow{0}

Theorems of the Cross Product

∣∣u→×v→∣∣=∣∣u→∣∣∗∣∣v→∣∣∗sinθu→×v→=0  ⟺  u→=cv→,c∈R∣∣u→×v→∣∣= The area of the parallelogram with u and v as adjacent sides|| \overrightarrow{u} \times \overrightarrow{v} || = || \overrightarrow{u} || * ||\overrightarrow{v} || * sin \theta \\ \overrightarrow{u} \times \overrightarrow{v} = 0 \iff \overrightarrow{u} = c \overrightarrow{v}, c \in \mathbb{R} \\ || \overrightarrow{u} \times \overrightarrow{v} || = \text{ The area of the parallelogram with u and v as adjacent sides}

The Triple Scalar Product

For u→,v→,w→∈R3, the triple scalar product is u→⋅(v→×w→)=det(u1u2u3v1v2v3w1w2w3)\text{For } \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \in \mathbb{R}^3 \text{, the triple scalar product is } \\ \overrightarrow{u} \cdot (\overrightarrow{v} \times \overrightarrow{w}) = det \begin{pmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{pmatrix}

The absolute value of the triple scalar product gives the volume of a parallelopiped where u, v, and w are adjacent edges.

3D Equations of a Line

A line L parallel to a vector v=<a,b,c>and passing through the point P(x1,y1,z1) is represented by the following parametric equations: x=x1+at,y=y1+bt,z=z1+cta, b, and c are the direction numbers, and t is the parameter. You can find any point on the line by changing t.\text{A line L parallel to a vector } v = <a, b, c> \text{and passing through the point } P(x_1, y_1, z_1) \text{ is represented by the following parametric equations: } \\ x = x_1 + at, y = y_1 + bt, z = z_1 + ct \\ \text{a, b, and c are the direction numbers, and t is the parameter. You can find any point on the line by changing t.}
Symmetric equations are the same as parametric except they are in terms of t.t=x−x1a,t=y−y1b,z−z1c\text{Symmetric equations are the same as parametric except they are in terms of t.} \\ t = \frac{x - x_1}{a}, t = \frac{y - y_1}{b}, \frac{z - z_1}{c}

Vector Normal to a Plane

Given a point P(x, y, z) and a line in symmetric/parametric form, the equation of the plane is given by the following equations:

General form: a(x−x1)+b(y−y1)+c(z−z1)=0Standard form: ax+by+cz=d where d=ax1+by1+cz1The normal vector to the plane is given by <a,b,c>\text{General form: } a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \\ \text{Standard form: } ax + by + cz = d \text{ where } d = ax_1 + by_1 + cz_1 \\ \text{The normal vector to the plane is given by } <a, b, c>

Angle Between Two Normal Vectors

To get the angle between two planes, you can use the angles between their two normal vectors

Distance Between a Point and a Plane

The distance between a plane and a point Q not in the plane is D=∣∣projn→PQ→∣∣=∣PQ→⋅n→∣∣∣n→∣∣where P is a point on the plane and n is a vector normal to the plane\text{The distance between a plane and a point Q not in the plane is } \\ D = || proj_{\overrightarrow{n}} \overrightarrow{PQ} || = \frac{|\overrightarrow{PQ} \cdot \overrightarrow{n} |}{|| \overrightarrow{n} ||} \\ \text{where P is a point on the plane and n is a vector normal to the plane}

Equations of Surfaces in Space

Ellipsoid: x2a2+y2b2+z2c2=1Hyperboloid of One Sheet: x2a2+y2b2−z2c2=1 (one of the terms must be negative, and the hyperbeloid goes along the axis made negative)Hyperboloid of Two Sheets: x2a2−y2b2−z2c2=1 (two of the terms must be negative, and the hyperbeloid goes along the axis made positive)Elliptic Cone: x2a2+y2b2−z2c2=0 (one of the terms must be negative, and the cone goes along the axis made negative)Elliptic Paraboloid: z=x2a2+y2b2 (the single variable denotes which direction it goes in; can be equal to x or y as well)Hyperbolic Paraboloid: z=x2a2−y2b2 (the single variable denotes which direction it goes in; can be equal to x or y as well)\text{Ellipsoid: } \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \\ \text{Hyperboloid of One Sheet: } \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \\ \text{ (one of the terms must be negative, and the hyperbeloid goes along the axis made negative)} \\ \text{Hyperboloid of Two Sheets: } \frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \\ \text{ (two of the terms must be negative, and the hyperbeloid goes along the axis made positive)} \\ \text{Elliptic Cone: } \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0 \\ \text{ (one of the terms must be negative, and the cone goes along the axis made negative)} \\ \text{Elliptic Paraboloid: } z = \frac{x^2}{a^2} + \frac{y^2}{b^2} \\ \text{ (the single variable denotes which direction it goes in; can be equal to x or y as well)} \\ \text{Hyperbolic Paraboloid: } z = \frac{x^2}{a^2} - \frac{y^2}{b^2} \\ \text{ (the single variable denotes which direction it goes in; can be equal to x or y as well)} \\

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Distance Between a Point and a Line

The distance between a line and a point P is D=∣∣PQ→×v→∣∣∣∣v→∣∣where v is a direction vector of the line and P is a point on the line\text{The distance between a line and a point P is } \\ D = \frac{|| \overrightarrow{PQ} \times \overrightarrow{v} ||}{|| \overrightarrow{v} ||} \\ \text{where v is a direction vector of the line and P is a point on the line}

Coordinate Conversion

Cylindrical to Rectangular: x=r⋅cosθ,y=r⋅sinθ,z=zRectangular to Cylindrical: r2=x2+y2,tanθ=yx,z=z\text{Cylindrical to Rectangular: } x = r \cdot cos \theta , y = r \cdot sin \theta , z = z \\ \text{Rectangular to Cylindrical: } r^2 = x^2 + y^2 , tan \theta = \frac{y}{x} , z = z \\
Spherical to Rectangular: x=ρ⋅sinϕ⋅cosθ,y=ρ⋅sinϕ⋅sinθ,z=ρ⋅cosϕRectangular to Spherical: ρ2=x2+y2+z2,tanθ=yx,ϕ=arccos(zx2+y2+z2)\text{Spherical to Rectangular: } x = \rho \cdot sin \phi \cdot cos \theta , y = \rho \cdot sin \phi \cdot sin \theta , z = \rho \cdot cos \phi \\ \text{Rectangular to Spherical: } \rho^2 = x^2 + y^2 + z^2 , tan \theta = \frac{y}{x} , \phi = arccos(\frac{z}{\sqrt{x^2 + y^2 + z^2}}) \\
Spherical to Cylindrical (r > 0): r2=ρ2⋅sin2ϕ,θ=θ,z=ρ⋅cosϕCylindrical to Spherical (r > 0): ρ=r2+z2,θ=θ,ϕ=arccos(zr2+z2)\text{Spherical to Cylindrical (r > 0): } r^2 = \rho^2 \cdot sin^2 \phi , \theta = \theta , z = \rho \cdot cos \phi \\ \text{Cylindrical to Spherical (r > 0): } \rho = \sqrt{r^2 + z^2} , \theta = \theta , \phi = arccos(\frac{z}{\sqrt{r^2 + z^2}}) \\