3D Coordinate

We use a tuple of three numbers to express a point in the 3D space as $(x,y,z)$. Note the right-hand rule for the three axis.

The distance between two points $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ is $$|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_1-y_2)^2+(z_1-z_2)^2}.$$

The Standard Equation for the Sphere of Radius $a$ and Center $(x_0,y_0,z_0)$ is $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=a^2.$$

Example

Find the center and radius of the sphere $$x^2+y^2+z^2+3x-4z+1=0.$$

Note that the equation is equivalent to $$(x+3/2)^2 + y^2 + (z-2)^2 = 9/4 + 4 - 1 = 21/4$$

Vector in 3D

Vectors represent things that have both magnitude and direction in the plane (2D) or in space (3D).

Vector/directed line segment represent quantities such as force, displacement, and velocity.

The vector represented by the directed line segment $\overrightarrow{AB}$ has initial point $A$ and terminal point $B$ and its length is denoted by $|\overrightarrow{AB}|$. Two vectors are equal if they have the same length (magnitude) and direction.

Component Form

If $\vec{v}$ is a two-dimensional vector in the plane equal to the vector with initial point at the origin and terminal point $(v_1,v_2)$, then the component form of $\vec{v}$ is $$\vec{v}=\langle v_1,v_2\rangle.$$

If $\vec{v}$ is a three-dimensional vector in the space equal to the vector with initial point at the origin and terminal point $(v_1,v_2,v_3)$, then the component form of $\vec{v}$ is $$\vec{v}=\langle v_1,v_2,v_3\rangle.$$

Given the points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the component form of $\overrightarrow{PQ}$ is $$\overrightarrow{PQ}=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle.$$

The magnitude or length of the vector $\vec{v}=\overrightarrow{PQ}$ is the nonnegative number $$|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.$$

Vector Algebra Operations

Let $\vec{u}=\langle u_1,u_2,u_3\rangle$ and $\vec{v}=\langle v_1,v_2,v_3\rangle$ be vectors in space and $k$ is a scalar in $R$.

• $\vec{u}+\vec{v}=\langle u_1+v_1,u_2+v_2,u_3+v_3\rangle$

• Scalar multiplication: $k\vec{u}=\langle ku_1,ku_2,ku_3\rangle$

Midpoint of a Line Segment

The midpoint $M$ of the line segment joining points $P(x_1,y_1,z_1)$ to $Q(x_2,y_2,z_2)$ is the point $$\left({x_1+x_2\over 2},{y_1+y_2\over 2},{z_1+z_2\over 2}\right).$$

Properties of Vector Operations

Let $\vec{u}$, $\vec{v}$, $\vec{w}$ be vectors in plane or space and $a$, $b$ are scalars.

• $\vec{u}+\vec{v}=\vec{v}+\vec{u}$

• $(\vec{u}+\vec{v})+\vec{w}=\vec{u}+(\vec{v}+\vec{w})$

• $\vec{u}+\vec{0}=\vec{0}+\vec{u}$

• $\vec{u}+(-\vec{u})=\vec{0}$

• $0\vec{u}=\vec{0}$

• $1\vec{u}=\vec{u}$

• $a(b\vec{u})=(ab)\vec{u}$

• $a(\vec{u}+\vec{v})=a\vec{u}+a\vec{v}$

• $(a+b)\vec{u}=a\vec{u}+b\vec{u}$

Unit Vectors/Direction

A vector $\vec{v}$ of length 1 is called a unit vector. The standard unit vectors are $$\vec{i}=\langle 1,0,0\rangle,~\vec{j}=\langle 0,1,0\rangle,~\vec{k}=\langle 0,0,1\rangle.$$ For a vector in space, we have $$\vec{v}={\color{red}?\quad}\vec{i}+{\color{red}?\quad}\vec{j}+{\color{red}?\quad}\vec{k}.$$

Find a unit vector $\vec{u}$ in the direction of the vector from $P(1,0,1)$ to $Q(3,2,0)$.

First we find the vector form $P$ to $Q$, which is $$\vec{v}=\langle 3-1, 2-0, 0-1\rangle=\langle 2, 2, -1\rangle.$$ Then we normalize it by dividing by the length of the vector ($|\vec{v}|=\sqrt{2^2+2^2+(-1)^2}=3$) to get the unit vector $${\vec{v}\over |\vec{v}|}={1\over 3}\langle 2, 2, -1\rangle=\left\langle {2\over 3}, {2\over 3}, -{1\over 3}\right\rangle.$$

Length and Direction

If $\vec{v}\neq \vec{0}$, then

• ${\vec{v}\over |\vec{v}|}$ is the direction of $\vec{v}.$

• the equation $\vec{v}=|\vec{v}|{\vec{v}\over |\vec{v}|}$ expresses $\vec{v}$ as its length times its direction.

The Dot Product

If a force $\vec{F}$ is applied to a particle moving along a straight path, we often need to know the magnitude of the force in the direction of motion. Note that the direction that is perpendicular to the path does not help moving the particle along the path, so we only need to find the part that is parallel to the path. If the angle between the force and the path direction is $\theta$, then only the magnitude $|\vec{F}|\cos(\theta)$ works.

If $\vec{v}$ is parallel to the tangent line to the path at the point where $\vec{F}$ is applied, then we want the magnitude of $\vec{F}$ in the direction of $\vec{v}$, which is $|\vec{F}|\cos\theta$ where $\theta$ is the angle between $\vec{F}$ and the $\vec{v}$. The question is "how do we find the angle $\theta$?"

By the cosine rule, we have $$|\vec{u}-\vec{v}|^2=|\vec{u}|^2+|\vec{v}|^2-2|\vec{u}||\vec{v}|\cos\theta.$$

Inner product of u and v: u*v, u.dot_product(v), u.inner_product(v)

Norm of u: u.norm()

Angle Between Two Vectors

The dot product (inner product) $\vec{u}\cdot\vec{v}$ of vectors $\vec{u}=\langle u_1,u_2,u_3\rangle$ and $\vec{v}=\langle v_1,v_2,v_3\rangle$ is the scalar $$\vec{u}\cdot\vec{v}={u_1v_1+u_2v_2+u_3v_3}.$$

So the cosine rule tells $$|\vec{u}-\vec{v}|^2= |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u}\cdot\vec{v}=|\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta.$$

The angle $\theta$ between two nonzero vectors $\vec{u}=\langle u_1,u_2,u_3\rangle$ and $\vec{v}=\langle v_1,v_2,v_3\rangle$ is given by $$\theta=\cos^{-1}\left({\vec{u}\cdot\vec{v}\over|\vec{u}||\vec{v}|}\right).$$

Example

Find the angle between $\vec{u}=\vec{i}-2\vec{j}-2\vec{k}$ and $\vec{v}=6\vec{i}+3\vec{j}+2\vec{k}$.

Find the angle $\theta$ in the triangle $ABC$ determined by the vertices $A = (0, 0),~B = (3, 5)$, and $C = (5, 2)$.

We first compute the two vectors $\vec{u}:=CA = (0-5,0-2)=(-5,-2)$, and $\vec{v}:=CB=(3-5,5-2)=(-2,3)$. Then we compute the angle by the formule $$\theta=\cos^{-1}\left({\vec{u}\cdot\vec{v}\over|\vec{u}||\vec{v}|}\right).$$

Orthogonal Vectors

Vectors $\vec{u}$ and $\vec{v}$ are orthogonal if $\vec{u}\cdot\vec{v}=0$.

Properties of the Dot Product

If $\vec{u}$, $\vec{v}$, and $\vec{w}$ are any vectors and $c$ is a scalar, then

• $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$

• $(c\vec{u})\cdot\vec{v}=\vec{u}\cdot(c\vec{v})=c(\vec{u}\cdot\vec{v})$

• $\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$

• $\vec{u}\cdot\vec{u}=|\vec{u}|^2$

• $\vec{0}\cdot\vec{u}=0$

Projecting one Vector onto another

The vector projection of $\vec{u}$ onto $\vec{v}$ is the vector $$\mbox{proj}_{\vec{v}}\vec{u}=(|\vec{u}|\cos\theta){\vec{v}\over |\vec{v}|}$$ The scalar component of $\vec{u}$ in the direction of $\vec{v}$ is the scalar $$|\vec{u}|\cos\theta = {\vec{u}\cdot\vec{v}\over |\vec{v}|}=\vec{u}\cdot{\vec{v}\over|\vec{v}|}$$

Therefore, the projection is $$\mbox{proj}_{\vec{v}}\vec{u}={\vec{u}\cdot\vec{v}\over|\vec{v}|^2}{\vec{v}}$$

Verify that $\vec{u}-\mbox{proj}_{\vec{v}}\vec{u}$ is orthogonal to $\vec{v}$.

$$(\vec{u}-\mbox{proj}_{\vec{v}}\vec{u})\cdot \vec{v}=\vec{u}\cdot\vec{v}-\mbox{proj}_{\vec{v}}\vec{u}\cdot \vec{v}=\vec{u}\cdot\vec{v}-{\vec{u}\cdot\vec{v}\over|\vec{v}|^2}{\vec{v}\cdot\vec{v}}=0.$$

The Cross Product

In a plane, we can use slope and angle of inclination to describe a line. How to describe a plane in space?

If $\vec{u}$ and $\vec{v}$ are not parallel, they determine a plane. We select a unit vector $\vec{n}$ perpendicular to the plane by the right-hand rule.

The cross/vector product $\vec{u}\times \vec{v}$ is the vector $$\vec{u}\times \vec{v}=(|\vec{u}||\vec{v}|\sin\theta)\vec{n}.$$

• $\vec{i}\times\vec{j}=\vec{k}=-(\vec{j}\times \vec{i})$

• $\vec{j}\times\vec{k}=\vec{i} = -(\vec{k}\times \vec{j})$

• $\vec{k}\times \vec{i}=\vec{j}= -(\vec{i}\times \vec{k})$

• $\vec{i}\times \vec{i}=\vec{j}\times \vec{j}=\vec{k}\times \vec{k}=\vec{0}$

Applications of the cross product:

• compute area The magnitude of the cross product is the area of the parallelogram determined by $\vec{u}$ and $\vec{v}$.

• find perpendicular vector: $\vec{u}\times \vec{v}$ is perpendicular to $\vec{u}$ and $\vec{v}$.

Nonzero vectors $\vec{u}$ and $\vec{v}$ are parallel if and only if $\vec{u}\times \vec{v}=\vec{0}$

Some Properties of the Cross Product

If $\vec{u}$, $\vec{v}$, and $\vec{w}$ are any vectors and $r,~s$ are scalars, then

• $(r\vec{u})\times(s\vec{v})=(rs)(\vec{u}\times\vec{v})$

• $\vec{u}\times \vec{v}=-\vec{v}\times \vec{u}$

• $\vec{0}\cdot\vec{u}=\vec{0}$

• $\vec{u}\times(\vec{v}+\vec{w})=\vec{u}\times\vec{v}+\vec{u}\times\vec{w}$

• $(\vec{v}+\vec{w})\times \vec{u}=\vec{v}\times\vec{u}+\vec{w}\times\vec{u}$

• $\vec{u}\times (\vec{v}\times \vec{w}) = (\vec{u}\cdot\vec{w})\vec{v}-(\vec{u}\cdot\vec{v})\vec{w}$

Determinant Formula for $\vec{u}\times \vec{v}$

If $\vec{u}=\langle u_1,u_2,u_3\rangle$ and $\vec{v}=\langle v_1,v_2,v_3\rangle$, then based on the properties of cross product, we have $$\vec{u}\times\vec{v}=(u_1\vec{i}+u_2\vec{j}+u_3\vec{k})\times(v_1\vec{i}+v_2\vec{j}+v_3\vec{k})={\color{blue}u_1v_2\vec{k}}{\color{red}-u_1v_3\vec{j}}{\color{blue}-u_2v_1\vec{k}}+{\color{green}u_2v_3\vec{i}}+{\color{red}u_3v_1\vec{j}}{\color{green}-u_3v_2\vec{i}}=\begin{vmatrix} {\color{green}\vec{i}} & {\color{red}\vec{j}} & {\color{blue}\vec{k}}\\u_1&u_2 & u_3\\v_1&v_2&v_3 \end{vmatrix}$$

Triple Scalar or Box Product

It computes the volume of the parallelepiped determined by $\vec{u}$, $\vec{v}$, and $\vec{w}$.

Calculating the Triple Scalar Product as a Determinant $$(\vec{u}\times\vec{v})\cdot\vec{w}=((u_1\vec{i}+u_2\vec{j}+u_3\vec{k})\times(v_1\vec{i}+v_2\vec{j}+v_3\vec{k}))\cdot(w_1\vec{i}+w_2\vec{j}+w_3\vec{k})={\color{green}(u_2v_3-u_3v_2)}w_1+{\color{red}(u_3v_1-u_1v_3)}w_2+{\color{blue}(u_1v_2-u_2v_1)}w_3=\begin{vmatrix}w_1&w_2&w_3\\ u_1&u_2 & u_3\\v_1&v_2&v_3 \end{vmatrix}=\begin{vmatrix} u_1&u_2 & u_3\\v_1&v_2&v_3\\w_1&w_2&w_3 \end{vmatrix}$$

Lines and Planes in Space

A vector equation for the line $L$ through $P_0(x_0,y_0,z_0)$ parallel to $\vec{v}$ is $$\vec{r}(t)=\vec{r}_0+t\vec{v},\qquad -\infty<t<\infty,$$ where $\vec{r}_0$ is the position vector of $P_0$.

The standard parametrization of the line through $P_0(x_0,y_0,z_0)$ parallel to $\vec{v}=v_1\vec{i}+v_2\vec{j}+v_3\vec{k}$ is $$x=x_0+tv_1, \quad y = y_0+tv_2,\quad z=z_0+tv_3,\quad -\infty<t<\infty.$$

Example: Parametrize the line segment joining the points $P(-3, 2, -3)$ and $Q(1,-1,4)$

We need to find $\vec{v}$ based on the two points on the line. By calculation, the vector $\vec{v}$ is $$\vec{v}=\langle 1-(-3), -1-2, 4-(-3)\rangle =\langle 4, -3,7\rangle$$ Then we have $$x = -3 + 4t,\quad y = 2 - 3t,\quad z = -3 + 7t,\quad 0\leq t \leq1$$

Distance from a Point $S$ to a Line Through $P$ Parallel to $\vec{v}$

The distance is from a point to a line is $|PS|\sin\theta$, which does not depend on the point $P$.

Find the distance from the point $S(1, 1, 5)$ to the line $$L: x = 1 + t,~y = 3 - t,~z = 2t.$$

An Equation for a Plane in Space

A plane is determined by a point on the plane and its "tilt" or orientation.

The plane through $P_0(x_0,y_0,z_0)$ normal to $\vec{n}=A\vec{i}+B\vec{j}+C\vec{k}=\langle A,B,C\rangle$ has

• Vector equation: $\vec{n}\cdot\vec{P_0P}=0$

• Component equation: $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$

• Component equation simplified: $Ax+By+Cz = Ax_0+By_0+Cz_0$

Example: Find an equation for the plane through $A(0,0,1)$, $B(2,0,0)$, and $C(0,3,0)$.

We need to find the normal vector first use the cross product of $\vec{AB}= 2\vec{i}-\vec{k}$ and $\vec{AC}=3\vec{j}-\vec{k}$. The cross product is $\vec{AB}\times \vec{AC}=3\vec{i}+2\vec{j}+6\vec{k}$. Therefore, the equation is $$3x+2y+6z = 6.$$