Vectors & the Geometry of Space
Drag the figure to rotate it
Three coordinates locate a point in space
To reach $P(a,b,c)$, walk $a$ along $x$, then $b$ along $y$, then $c$ along $z$.
Right-hand rule
Curl the fingers of your right hand from $+x$ toward $+y$. Your thumb points along $+z$.
- Coordinate planes
- $xy$: $z=0$ · $yz$: $x=0$ · $xz$: $y=0$
- Octants
- The three planes cut space into 8 octants. In the first octant, $x,y,z>0$.
- Projections of $P$
- $(a,b,0)$, $(0,b,c)$, $(a,0,c)$: drop $P$ straight onto each plane.
Drag the figure to rotate it: a flat picture cannot show depth.
In ℝ³, one equation describes a surface
A variable missing from the equation is free. Draw the curve in the other two variables, then slide it along the missing axis.
| Equation | In ℝ² | In ℝ³ |
|---|---|---|
| $x=2$ | line | plane parallel to the $yz$-plane |
| $y=x$ | line | vertical plane |
| $x^2+y^2=4$ | circle | cylinder around the $z$-axis |
| $z=y^2$ | parabola | parabolic cylinder |
- Two equations
- usually a curve: $x^2+y^2=4,\ z=3$ is a circle at height 3
- Inequalities
- a solid: $1\le x^2+y^2+z^2\le4$ is a thick spherical shell
Drag the figure to rotate it
Distance in space, and the sphere it defines
Distance comes from the Pythagorean theorem, used twice. A sphere is every point at distance $r$ from a center $C$.
- 1. Floor
- diagonal $d$ of the floor: $\ d^2=\Delta x^2+\Delta y^2$
- 2. Up
- $|P_1P_2|^2=d^2+\Delta z^2=\Delta x^2+\Delta y^2+\Delta z^2$
This is the sphere with center $C(h,k,l)$ and radius $r$. From the expanded form, complete the square in $x$, $y$, and $z$ to find $C$ and $r$.
Example: find the center and radius
Center $(-2,3,-1)$, radius $\sqrt8=2\sqrt2$.
A negative right side means no points; $0$ means only the center.
Legs Δx, Δy, Δz · dashed: floor diagonal $d$ · magenta: $CP$ · drag to rotate
A vector is a displacement: add it, subtract it, scale it
A vector has a length and a direction, but no fixed position. Sliding an arrow without turning or stretching it gives the same vector.
- $\mathbf a+\mathbf b$
- Add components. Picture: put the tail of $\mathbf b$ at the head of $\mathbf a$.
- $\mathbf a-\mathbf b$
- $\mathbf a+(-\mathbf b)$: the arrow from the head of $\mathbf b$ to the head of $\mathbf a$.
- $c\,\mathbf a$
- $|c|$ times as long as $\mathbf a$, and reversed when $c<0$.
- Parallel
- $\mathbf b$ is parallel to $\mathbf a$ ($\mathbf a\parallel\mathbf b$) if $\mathbf b=c\,\mathbf a$ ($c\ne0$). If $\mathbf b$ starts from a point $P$, then it is $P+c\,\mathbf a$.
Choose an operation to animate it · drag a and b (or point P)
Length and unit vectors
Divide a nonzero vector by its length. The result keeps the direction and has length 1.
- Basis
- $\mathbf i=\langle1,0,0\rangle$, $\mathbf j=\langle0,1,0\rangle$, $\mathbf k=\langle0,0,1\rangle$, so $\langle a_1,a_2,a_3\rangle=a_1\mathbf i+a_2\mathbf j+a_3\mathbf k$
- Length $L$
- $L\,\mathbf u$ points along $\mathbf a$; $\ -\mathbf u$ points the opposite way
- Example
- $\mathbf a=2\mathbf i-\mathbf j-2\mathbf k$: $\ |\mathbf a|=\sqrt{4+1+4}=3$, $\ \mathbf u=\left\langle\tfrac23,-\tfrac13,-\tfrac23\right\rangle$
Drag the head of a: the head of u always lands on the unit circle (the unit sphere in 3D).
The dot product turns two vectors into a number
Multiply matching components and add. The sign of the result tells you whether the angle is acute, right, or obtuse.
| $\mathbf a\cdot\mathbf b>0$ | $0\le\theta<\pi/2$ — acute |
| $\mathbf a\cdot\mathbf b=0$ | $\theta=\pi/2$ — orthogonal |
| $\mathbf a\cdot\mathbf b<0$ | $\pi/2<\theta\le\pi$ — obtuse |
- Rules
- $\mathbf a\cdot\mathbf a=|\mathbf a|^2,\quad \mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a,$
$\mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c,\quad (c\mathbf a)\cdot\mathbf b=c(\mathbf a\cdot\mathbf b)$
Dashed violet line: every direction ⊥ a · drag the heads of a and b
Find the angle between two vectors
Solve $\mathbf a\cdot\mathbf b=|\mathbf a|\,|\mathbf b|\cos\theta$ for $\cos\theta$.
Example
$\mathbf a=\langle2,2,-1\rangle$, $\mathbf b=\langle5,-3,2\rangle$: $\ \mathbf a\cdot\mathbf b=10-6-2=2$, $\ |\mathbf a|=3$, $\ |\mathbf b|=\sqrt{38}$
$\theta=\arccos\dfrac{2}{3\sqrt{38}}\approx84^\circ$
Orthogonal means $\mathbf a\cdot\mathbf b=0$. Example: $\langle2,2,-1\rangle\cdot\langle5,-4,2\rangle=10-8-2=0$.
Why is $\mathbf a\cdot\mathbf b=|\mathbf a|\,|\mathbf b|\cos\theta$?
Compute $|\mathbf a-\mathbf b|^2$ for the triangle on the right in two ways:
Both equal $|\mathbf a-\mathbf b|^2$, so $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$.
$\arccos$ returns an angle in $[0,\pi]$, the smaller angle between the vectors. Leave the answer exact unless a decimal is asked for.
Drag the heads of a and b: both ways of computing |a − b|² always agree.
Projection: the shadow of b on a
Shine light perpendicular to $\mathbf a$. The shadow that $\mathbf b$ casts on the line through $\mathbf a$ is the projection.
Magenta: projection · teal: orthogonal part · drag the heads of a and b
A line through a point, and where it meets a sphere
Start at a point and add multiples of a direction $\mathbf v$. Each step of 1 in $t$ moves a distance $|\mathbf v|$.
$\mathbf r_0$: position vector of a point on the line · $\mathbf v$: its direction. (Preview of §12.5.)
Through the center $C$ of a sphere of radius $r$
So the line meets the sphere at $C\pm r\,\dfrac{\mathbf v}{|\mathbf v|}$: walk $r$ along the unit vector, in both directions.
Line not through the center? Substitute $x(t),y(t),z(t)$ into the sphere’s equation and solve the quadratic in $t$: 2, 1, or 0 solutions.
$C(2,1,1)$, $r=6$, $\mathbf v=\langle2,-1,2\rangle$. Dots mark whole steps of $t$, each $|\mathbf v|=3$ long, so the line exits at $t=\pm2$.
Practice: spheres, lines, and dot products
Try each problem before looking at the solutions on the next slides. Give exact answers.
Exercise 1 · Find the sphere
- Show that this is a sphere. Find its center $C$ and radius $r$.
- Describe where the sphere meets the $xy$-plane.
- Is the origin inside, on, or outside the sphere?
Exercise 2 · Line meeting a sphere
Use the sphere $S$ from Exercise 1 and the point $E(9,2,-4)$.
- Write a vector equation of the line $L$ through $C$ and $E$.
- Find the exact points where $L$ meets $S$.
Exercise 3 · A triangle in space
$A(1,0,2),\ B(3,4,6),\ D(3,-1,4)$
- Find the unit vector that points in the direction opposite to $\overrightarrow{AD}$.
- Find the exact angle $\angle BAD$.
- Find the scalar and vector projections of $\overrightarrow{AD}$ onto $\overrightarrow{AB}$.
- Write $\overrightarrow{AD}$ as the sum of a vector parallel to $\overrightarrow{AB}$ and a vector orthogonal to $\overrightarrow{AB}$.
Solution 1: complete the square to find the sphere
Group the terms by variable, complete each square, and add the same constants to the right side.
(a) Complete the square in $x$, $y$, and $z$:
(b) The $xy$-plane is $z=0$:
(c) Compare $|OC|$ with $r$:
- (a) Center $C(3,-1,2)$, radius $r=6$
- (b) A circle in the $xy$-plane with center $(3,-1,0)$ and radius $\sqrt{32}=4\sqrt2$
- (c) Inside, because $\sqrt{14}<6$
Solution 2: where the line meets the sphere
The line passes through the center, so the parameter $t$ measures distance from $C$ in steps of $|\mathbf v|$.
(a) $\overrightarrow{CE}=\langle6,3,-6\rangle=3\langle2,1,-2\rangle$. Use the simpler direction vector:
(b) $|\mathbf r(t)-C|=|t|\,|\langle2,1,-2\rangle|=3|t|$. Setting $3|t|=6$ gives $t=\pm2$:
Check by substitution: $(2t)^2+t^2+(-2t)^2=9t^2=36$, so $t=\pm2$.
- (a) $\mathbf r(t)=\langle3,-1,2\rangle+t\langle2,1,-2\rangle$ (any nonzero multiple of the direction works)
- (b) $(7,1,-2)$ and $(-1,-3,6)$
Solution 3: angle and projection in a triangle
Find the two edge vectors from $A$ first. Every part then uses the same dot product.
(a) $-\dfrac{\overrightarrow{AD}}{|\overrightarrow{AD}|}=\left\langle-\tfrac23,\ \tfrac13,\ -\tfrac23\right\rangle$
(b) $\overrightarrow{AB}\cdot\overrightarrow{AD}=4-4+8=8$, so
(c) Divide by the length of $\overrightarrow{AB}$, the vector you are projecting onto:
(d) Subtract the projection to get the orthogonal part:
Check: $\left\langle\tfrac{14}9,-\tfrac{17}9,\tfrac{10}9\right\rangle\cdot\langle2,4,4\rangle=\tfrac{28-68+40}{9}=0$ ✓
Key formulas and what they mean
Learn each formula together with a picture of what it measures.
| Idea | Formula | Picture it as |
|---|---|---|
| Distance | $\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$ | The diagonal of a box |
| Sphere | $(x-h)^2+(y-k)^2+(z-l)^2=r^2$ | All points at distance $r$ from $C$. Complete the square to find $C$ and $r$. |
| Vector $\overrightarrow{AB}$ | $\langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\rangle$ | Head minus tail |
| Unit vector | $\mathbf a/|\mathbf a|$ | Same direction as $\mathbf a$, length 1 |
| Dot product | $a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta$ | Its sign tells you whether $\theta$ is acute, right, or obtuse |
| Projection | $\operatorname{comp}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|}$, $\ \operatorname{proj}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|^2}\mathbf a$ | The shadow of $\mathbf b$ on the line through $\mathbf a$ |
| Line | $\mathbf r(t)=\mathbf r_0+t\mathbf v$ | Through a sphere's center, it meets the sphere at $t=\pm r/|\mathbf v|$ |