Ch.2: Vectors
Chapter 2 of the book "No Bullshit Guide to Linear Algebra" by Ivan Savov Basis Vector products Complex numbers
59 flashcards · Mathematics · by lakesare
Part of Textbooks › No Bullshit Guide to Linear Algebra (Ivan Savov) · 10 courses
[Vectors] What is a vector v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx,vy) , and what are its components?
An ordered list of numbers; those numbers ( vxv_xvx , vyv_yvy ) are its components. Geometrically, v⃗\vec{v}v is the arrow that starts at the origin (0,0)(0, 0)(0,0) and ends at the point (vx,vy)(v_x, v_y)(vx,vy) . Example: v⃗=(3,2)\vec{v} = (3, 2)v=(3,2) is the arrow from (0,0)(0, 0)(0,0) to (3,2)(3, 2)(3,2) , with xxx -component 3 and yyy -component 2.
[Vectors] The word vector comes from the Latin vehere, which means to carry.
The vector v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx,vy) takes the point (0,0)(0, 0)(0,0) to the point (vx,vy)(v_x, v_y)(vx,vy) .
[Vectors] A tennis ball has velocity v⃗1=(20,0)\vec{v}_1 = (20, 0)v1=(20,0) or v⃗2=(0,20)\vec{v}_2 = (0, 20)v2=(0,20) (in m/s). What do the two have in common, and what differs?
Same speed (20 m/s), different directions, so they are different velocities. A velocity vector contains both the speed and the direction of motion; a single number gives only the speed.
[Vectors] What are the three notations for a vector in R2\mathbb{R}^2R2 ?
Component notation v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx,vy) : coordinates with respect to the xxx -axis and the yyy -axis. Example: (1,1)(1, 1)(1,1) Unit vector notation v⃗=vxı^+vyȷ^\vec{v} = v_x \hat{\imath} + v_y \hat{\jmath}v=vx^+vy^ , where ı^=(1,0)\hat{\imath} = (1, 0)^=(1,0) and ȷ^=(0,1)\hat{\jmath} = (0, 1)^=(0,1) . Example: ı^+ȷ^\hat{\imath} + \hat{\jmath}^+^ Length-and-direction notation v⃗=∥v⃗∥∠θ\vec{v} = \|\vec{v}\| \angle \thetav=∥v∥∠θ : the length ∥v⃗∥\|\vec{v}\|∥v∥ and…
[Vectors] For u⃗=(ux,uy)\vec{u} = (u_x, u_y)u=(ux,uy) , v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx,vy) and a number α∈R\alpha \in \mathbb{R}α∈R , how are addition, subtraction and scaling computed?
Addition u⃗+v⃗=(ux+vx, uy+vy)\vec{u} + \vec{v} = (u_x + v_x, \ u_y + v_y)u+v=(ux+vx, uy+vy) Example: (4,2)+(3,7)=(7,9)(4, 2) + (3, 7) = (7, 9)(4,2)+(3,7)=(7,9) Subtraction u⃗−v⃗=(ux−vx, uy−vy)\vec{u} - \vec{v} = (u_x - v_x, \ u_y - v_y)u−v=(ux−vx, uy−vy) Example: (4,2)−(3,7)=(1,−5)(4, 2) - (3, 7) = (1, -5)(4,2)−(3,7)=(1,−5) Scaling αu⃗=(αux, αuy)\alpha \vec{u} = (\alpha u_x, \ \alpha u_y)αu=(αux, αuy) Example: 3⋅(4,2)=(12,6)3 \cdot (4, 2) = (12, 6)3⋅(4,2)=(12,6) All three act on each…
[Vectors] What does scaling a vector v⃗\vec{v}v by a number α\alphaα do to v⃗\vec{v}v , depending on α\alphaα ?
α>1\alpha > 1α>1 αv⃗\alpha \vec{v}αv is longer than v⃗\vec{v}v , same direction. Example: 2⋅(1,0)=(2,0)2 \cdot (1, 0) = (2, 0)2⋅(1,0)=(2,0) 0≤α<10 \le \alpha < 10≤α<1 αv⃗\alpha \vec{v}αv is shorter than v⃗\vec{v}v , same direction. Example: 0.5⋅(3,2)=(1.5,1)0.5 \cdot (3, 2) = (1.5, 1)0.5⋅(3,2)=(1.5,1) α<0\alpha < 0α<0 αv⃗\alpha \vec{v}αv points in the opposite direction. Example: −1⋅(3,0)=(−3,0)-1 \cdot (3, 0) = (-3, 0)−1⋅(3,0)=(−3,0) , same length as (3,0)(3, 0)(3,0) .
[Vectors] What is the length ∥v⃗∥\|\vec{v}\|∥v∥ of v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx,vy) ?
∥v⃗∥=vx2+vy2\boldsymbol{\|\vec{v}\| = \sqrt{v_x^2 + v_y^2}}∥v∥=vx2+vy2 By Pythagoras' theorem: v⃗\vec{v}v is the hypotenuse of a right triangle with legs vxv_xvx and vyv_yvy . The length is also written v⃗⋅v⃗\sqrt{\vec{v} \cdot \vec{v}}v⋅v , or just vvv .Example: ∥(3,4)∥=9+16=5\|(3, 4)\| = \sqrt{9 + 16} = 5∥(3,4)∥=9+16=5 .
[Vectors] How do you get the unit vector v^\hat{v}v^ in the direction of a vector v⃗≠0⃗\vec{v} \neq \vec{0}v=0 ?
v^=v⃗∥v⃗∥\boldsymbol{\hat{v} = \dfrac{\vec{v}}{\|\vec{v}\|}}v^=∥v∥v , i.e. scale v⃗\vec{v}v by one over its length. ∥v^∥=1\|\hat{v}\| = 1∥v^∥=1 . A unit vector (written with a hat) describes a direction without a specific length.Example: v⃗=(3,4)\vec{v} = (3, 4)v=(3,4) , ∥v⃗∥=5\|\vec{v}\| = 5∥v∥=5 , so v^=(0.6,0.8)\hat{v} = (0.6, 0.8)v^=(0.6,0.8) .
[Vectors] How do you add two vectors v⃗1\vec{v}_1v1 and v⃗2\vec{v}_2v2 geometrically, as arrows?
Draw v⃗2\vec{v}_2v2 starting at the tip of v⃗1\vec{v}_1v1 ; the sum is the arrow from the tail of v⃗1\vec{v}_1v1 to the tip of v⃗2\vec{v}_2v2 . The sum is the combined displacement of the two vectors. Example: (3,0)+(2,2)=(5,2)(3, 0) + (2, 2) = (5, 2)(3,0)+(2,2)=(5,2) .
[Vectors] How do you compute w⃗−v⃗1\vec{w} - \vec{v}_1w−v1 geometrically, as arrows?
Add −v⃗1-\vec{v}_1−v1 to w⃗\vec{w}w : −v⃗1-\vec{v}_1−v1 has the same length as v⃗1\vec{v}_1v1 and the opposite direction. Example: (5,2)−(3,0)=(5,2)+(−3,0)=(2,2)(5, 2) - (3, 0) = (5, 2) + (-3, 0) = (2, 2)(5,2)−(3,0)=(5,2)+(−3,0)=(2,2) .
[Vectors] A vector in length-and-direction notation is ∥r⃗∥∠θ\|\vec{r}\| \angle \theta∥r∥∠θ . What are its components (rx,ry)(r_x, r_y)(rx,ry) ?
rx=∥r⃗∥cosθ,ry=∥r⃗∥sinθ\boldsymbol{r_x = \|\vec{r}\| \cos\theta, \quad r_y = \|\vec{r}\| \sin\theta}rx=∥r∥cosθ,ry=∥r∥sinθ Example: 2∠30∘=(2cos30∘, 2sin30∘)=(3, 1)2 \angle 30^\circ = (2 \cos 30^\circ, \ 2 \sin 30^\circ) = (\sqrt{3}, \ 1)2∠30∘=(2cos30∘, 2sin30∘)=(3, 1) .
[Vectors] A vector has components (rx,ry)(r_x, r_y)(rx,ry) with rx>0r_x > 0rx>0 . What is it in length-and-direction notation ∥r⃗∥∠θ\|\vec{r}\| \angle \theta∥r∥∠θ ?
∥r⃗∥=rx2+ry2,θ=tan−1 (ryrx)\boldsymbol{\|\vec{r}\| = \sqrt{r_x^2 + r_y^2}, \quad \theta = \tan^{-1}\!\left(\frac{r_y}{r_x}\right)}∥r∥=rx2+ry2,θ=tan−1(rxry) Example: (1,1)=2∠45∘(1, 1) = \sqrt{2} \angle 45^\circ(1,1)=2∠45∘ .
[Vectors] Why can θ=tan−1(ry/rx)\theta = \tan^{-1}(r_y / r_x)θ=tan−1(ry/rx) give the wrong direction for a vector (rx,ry)(r_x, r_y)(rx,ry) , and how is it fixed?
tan−1\tan^{-1}tan−1 only returns angles between −90∘-90^\circ−90∘ and 90∘90^\circ90∘ ; when rx<0r_x < 0rx<0 , add 180∘180^\circ180∘ . Example: (−1,−1)(-1, -1)(−1,−1) : tan−1(1)=45∘\tan^{-1}(1) = 45^\circtan−1(1)=45∘ , but the vector points down-left, so θ=45∘+180∘=225∘\theta = 45^\circ + 180^\circ = 225^\circθ=45∘+180∘=225∘ .
[Vectors] What is component notation good for, and what is length-and-direction notation good for?
Components: arithmetic (adding, subtracting). Length and direction: seeing the size of a vector. So a sum of vectors given as ∥v⃗∥∠θ\|\vec{v}\| \angle \theta∥v∥∠θ is computed by converting to components first.
[Vectors] What are the unit vectors ı^\hat{\imath}^ , ȷ^\hat{\jmath}^ , k^\hat{k}k^ , and how is (vx,vy,vz)(v_x, v_y, v_z)(vx,vy,vz) written in unit vector notation?
ı^=(1,0,0), ȷ^=(0,1,0), k^=(0,0,1)\boldsymbol{\hat{\imath} = (1, 0, 0), \ \hat{\jmath} = (0, 1, 0), \ \hat{k} = (0, 0, 1)}^=(1,0,0), ^=(0,1,0), k^=(0,0,1) , and (vx,vy,vz)=vxı^+vyȷ^+vzk^(v_x, v_y, v_z) = v_x \hat{\imath} + v_y \hat{\jmath} + v_z \hat{k}(vx,vy,vz)=vx^+vy^+vzk^ . They are the unit vectors in the xxx , yyy and zzz directions. Addition adds the ı^\hat{\imath}^ terms together and the ȷ^\hat{\jmath}^ terms together.Example: (2ı^+3ȷ^)+(5ı^−2ȷ^)=7ı^+1ȷ^(2\hat{\imath} + 3…
[Vectors] Compute s⃗=4ı^+5∠30∘\vec{s} = 4\hat{\imath} + 5 \angle 30^\circs=4^+5∠30∘ and give the answer in length-and-direction notation.
s⃗≈8.697∠16.7∘\boldsymbol{\vec{s} \approx 8.697 \angle 16.7^\circ}s≈8.697∠16.7∘ Convert to components first: 5∠30∘=5cos30∘ ı^+5sin30∘ ȷ^=532ı^+52ȷ^5 \angle 30^\circ = 5 \cos 30^\circ \, \hat{\imath} + 5 \sin 30^\circ \, \hat{\jmath} = \frac{5\sqrt{3}}{2} \hat{\imath} + \frac{5}{2} \hat{\jmath}5∠30∘=5cos30∘^+5sin30∘^=253^+25^ .So s⃗=(4+532)ı^+52ȷ^≈8.330ı^+2.5ȷ^\vec{s} = (4 + \frac{5\sqrt{3}}{2}) \hat{\imath} + \frac{5}{2} \hat{\jmath} \approx 8.330 \hat{\imath} + 2.5 \hat{\jmath}s=(4+…
[Vectors] Three forces act on a block: W⃗=300∠−90∘\vec{W} = 300 \angle {-90^\circ}W=300∠−90∘ , N⃗=260∠120∘\vec{N} = 260 \angle 120^\circN=260∠120∘ and F⃗f=50∠30∘\vec{F}_f = 50 \angle 30^\circFf=50∠30∘ . What is the net force F⃗net=W⃗+N⃗+F⃗f\vec{F}_{\text{net}} = \vec{W} + \vec{N} + \vec{F}_fFnet=W+N+Ff ?
F⃗net≈(−86.7, −49.8)≈100∠209.9∘\boldsymbol{\vec{F}_{\text{net}} \approx (-86.7, \ -49.8) \approx 100 \angle 209.9^\circ}Fnet≈(−86.7, −49.8)≈100∠209.9∘ Fnet,x=300cos(−90∘)+260cos120∘+50cos30∘=0−130+43.3=−86.7F_{\text{net},x} = 300 \cos(-90^\circ) + 260 \cos 120^\circ + 50 \cos 30^\circ = 0 - 130 + 43.3 = -86.7Fnet,x=300cos(−90∘)+260cos120∘+50cos30∘=0−130+43.3=−86.7 Fnet,y=300sin(−90∘)+260sin120∘+50sin30∘=−300+225.2+25=−49.8F_{\text{net},y} = 300 \sin(-90^\circ) + 260 \sin 120^\circ + 50…
[Vectors] A boat goes 12 knots relative to the water. The river current is 5 knots. What is the boat's speed relative to the shore when it goes straight upstream?
7 knots: 12ı^−5ı^=7ı^12\hat{\imath} - 5\hat{\imath} = 7\hat{\imath}12^−5^=7^ . The velocity relative to the shore is the sum of the boat's velocity relative to the water and the current's velocity.
[Vectors] A boat goes 12 knots relative to the water; the river current is 5 knots. At what angle θ\thetaθ against the current (measured from the line straight across) must it steer to cross the river perpendicularly, and how fast does it then cross?
θ=sin−1(5/12)≈24.62∘\boldsymbol{\theta = \sin^{-1}(5/12) \approx 24.62^\circ}θ=sin−1(5/12)≈24.62∘ , crossing speed 122−52≈10.91\sqrt{12^2 - 5^2} \approx 10.91122−52≈10.91 knots. The upstream component of the boat's velocity must cancel the current: 12sinθ=512 \sin\theta = 512sinθ=5 . The remaining component, 12cosθ12 \cos\theta12cosθ , points across. The hypotenuse 12 is the boat's velocity relative to the water; the short leg 5 cancels the current.
[Vectors] What is an nnn -dimensional vector?
A list of nnn numbers v⃗=(v1,v2,…,vn)∈Rn\vec{v} = (v_1, v_2, \dots, v_n) \in \mathbb{R}^nv=(v1,v2,…,vn)∈Rn . The rules of vector algebra are the same in every dimension; only drawing stops at three dimensions.Example: (1,0,−2,5)∈R4(1, 0, -2, 5) \in \mathbb{R}^4(1,0,−2,5)∈R4 .
[Vectors] In the xyzxyzxyz -coordinate system, what does the vector v⃗=3ı^+4ȷ^+2k^=(3,4,2)\vec{v} = 3\hat{\imath} + 4\hat{\jmath} + 2\hat{k} = (3, 4, 2)v=3^+4^+2k^=(3,4,2) describe?
Start at the origin (0,0,0)(0, 0, 0)(0,0,0) , move 3 units in the xxx -direction, 4 in the yyy -direction and 2 in the zzz -direction. The numbers are distances relative to the three perpendicular axes xxx , yyy , zzz . Other coordinate systems (bases) give other numbers for the same vector.
[Basis] What is a basis {e^1,e^2,e^3}\{\hat{e}_1, \hat{e}_2, \hat{e}_3\}{e^1,e^2,e^3} for R3\mathbb{R}^3R3 ?
A set of vectors used as a coordinate system: every v⃗∈R3\vec{v} \in \mathbb{R}^3v∈R3 can be written as v⃗=v1e^1+v2e^2+v3e^3\vec{v} = v_1 \hat{e}_1 + v_2 \hat{e}_2 + v_3 \hat{e}_3v=v1e^1+v2e^2+v3e^3 . The numbers (v1,v2,v3)(v_1, v_2, v_3)(v1,v2,v3) are the coefficients (coordinates) of v⃗\vec{v}v with respect to that basis: a distance v1v_1v1 in the e^1\hat{e}_1e^1 direction, v2v_2v2 in the e^2\hat{e}_2e^2 direction and v3v_3v3 in the e^3\hat{e}_3e^3 direction.Example: in the bas…
[Basis] The basis {ı^,ȷ^,k^}\{\hat{\imath}, \hat{\jmath}, \hat{k}\}{^,^,k^} of R3\mathbb{R}^3R3 , which belongs to the xyzxyzxyz -coordinate system, is called the standard basis.
Example: v⃗=(vx,vy,vz)=vxı^+vyȷ^+vzk^\vec{v} = (v_x, v_y, v_z) = v_x \hat{\imath} + v_y \hat{\jmath} + v_z \hat{k}v=(vx,vy,vz)=vx^+vy^+vzk^ .
[Basis] Colours can be written as coefficients in the RGB basis (red, green, blue) or in the CMY basis (cyan, magenta, yellow). Why does the triplet (33,66,99)(33, 66, 99)(33,66,99) alone not say which colour is meant?
The same triplet means different colours in different bases, so coefficients mean something only together with their basis. (33,66,99)RGB=33R+66G+99B(33, 66, 99)_{RGB} = 33\mathbf{R} + 66\mathbf{G} + 99\mathbf{B}(33,66,99)RGB=33R+66G+99B is "CoolBlue", and the same colour is (222,189,156)CMY(222, 189, 156)_{CMY}(222,189,156)CMY . Read in the CMY basis, (33,66,99)CMY(33, 66, 99)_{CMY}(33,66,99)CMY is a different colour.A subscript after the bracket names the basis of the coefficients.
[Basis] What does Savov say is the first thing to do when solving a problem with vectors?
Draw a coordinate system. The coordinate system (basis) connects the real-world vector quantities with their components; components are distances measured along that coordinate system.
[Vector products] Which two products of vectors are used in practice, and what kind of output does each give?
Dot product: a number. Cross product: a vector. ⋅:R3×R3→R\cdot : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}⋅:R3×R3→R and ×:R3×R3→R3\times : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3×:R3×R3→R3 . The componentwise product (vxwx,vywy,vzwz)(v_x w_x, v_y w_y, v_z w_z)(vxwx,vywy,vzwz) is not used.
[Vector products] What is the algebraic formula for the dot product v⃗⋅w⃗\vec{v} \cdot \vec{w}v⋅w of v⃗,w⃗∈R3\vec{v}, \vec{w} \in \mathbb{R}^3v,w∈R3 ?
v⃗⋅w⃗=vxwx+vywy+vzwz\boldsymbol{\vec{v} \cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z}v⋅w=vxwx+vywy+vzwz Example: (1,2,3)⋅(4,−5,6)=4−10+18=12(1, 2, 3) \cdot (4, -5, 6) = 4 - 10 + 18 = 12(1,2,3)⋅(4,−5,6)=4−10+18=12 .
[Vector products] What is the geometric formula for the dot product v⃗⋅w⃗\vec{v} \cdot \vec{w}v⋅w ?
v⃗⋅w⃗=∥v⃗∥∥w⃗∥cosφ\boldsymbol{\vec{v} \cdot \vec{w} = \|\vec{v}\| \|\vec{w}\| \cos\varphi}v⋅w=∥v∥∥w∥cosφ , where φ\varphiφ is the angle between v⃗\vec{v}v and w⃗\vec{w}w . Example: v⃗=(2,0)\vec{v} = (2, 0)v=(2,0) , w⃗=(3,3)\vec{w} = (3, 3)w=(3,3) : 2⋅32⋅cos45∘=6=2⋅3+0⋅3\ 2 \cdot 3\sqrt{2} \cdot \cos 45^\circ = 6 = 2 \cdot 3 + 0 \cdot 3 2⋅32⋅cos45∘=6=2⋅3+0⋅3 .
[Vector products] The dot product is also called the inner product or the scalar product.
"Scalar" because the result is a number that does not change when the basis changes.
[Vector products] How do you find the angle φ\varphiφ between two vectors v⃗\vec{v}v and w⃗\vec{w}w from their components?
φ=cos−1 (v⃗⋅w⃗∥v⃗∥∥w⃗∥)\boldsymbol{\varphi = \cos^{-1}\!\left(\dfrac{\vec{v} \cdot \vec{w}}{\|\vec{v}\| \|\vec{w}\|}\right)}φ=cos−1(∥v∥∥w∥v⋅w) Set the algebraic formula vxwx+vywy+vzwzv_x w_x + v_y w_y + v_z w_zvxwx+vywy+vzwz equal to the geometric formula ∥v⃗∥∥w⃗∥cosφ\|\vec{v}\| \|\vec{w}\| \cos\varphi∥v∥∥w∥cosφ and solve for φ\varphiφ .Example: (1,0,0)(1, 0, 0)(1,0,0) and (1,1,0)(1, 1, 0)(1,1,0) : cosφ=11⋅2\cos\varphi = \frac{1}{1 \cdot \sqrt{2}}cosφ=1⋅21 , so φ=45∘\varphi = 45^\c…
[Vector products] What is v⃗⋅w⃗\vec{v} \cdot \vec{w}v⋅w when v⃗\vec{v}v and w⃗\vec{w}w point in the same direction, are perpendicular, or point in opposite directions?
Same direction cos0∘=1\cos 0^\circ = 1cos0∘=1 , so v⃗⋅w⃗=∥v⃗∥∥w⃗∥\vec{v} \cdot \vec{w} = \|\vec{v}\| \|\vec{w}\|v⋅w=∥v∥∥w∥ . Example: (2,0)⋅(3,0)=6(2, 0) \cdot (3, 0) = 6(2,0)⋅(3,0)=6 Perpendicular cos90∘=0\cos 90^\circ = 0cos90∘=0 , so v⃗⋅w⃗=0\vec{v} \cdot \vec{w} = 0v⋅w=0 . Example: (2,0)⋅(0,3)=0(2, 0) \cdot (0, 3) = 0(2,0)⋅(0,3)=0 Opposite directions cos180∘=−1\cos 180^\circ = -1cos180∘=−1 , so v⃗⋅w⃗=−∥v⃗∥∥w⃗∥\vec{v} \cdot \vec{w} = -\|\vec{v}\| \|\vec{w}\|v⋅w=−∥v∥∥w∥ . Example: (2,0)⋅(−3…
[Vector products] What is the direction of the cross product a⃗×b⃗\vec{a} \times \vec{b}a×b , and for which vectors is it defined?
Perpendicular to both a⃗\vec{a}a and b⃗\vec{b}b ; defined only for vectors in R3\mathbb{R}^3R3 . Of the two perpendicular directions, the right-hand rule picks one.Example: a vector in the xxx -direction times a vector in the yyy -direction gives a vector in the zzz -direction.
[Vector products] What are the cross products ı^×ȷ^\hat{\imath} \times \hat{\jmath}^×^ , ȷ^×k^\ \hat{\jmath} \times \hat{k} ^×k^ and k^×ı^\hat{k} \times \hat{\imath}k^×^ of the standard basis vectors?
ı^×ȷ^=k^,ȷ^×k^=ı^,k^×ı^=ȷ^\boldsymbol{\hat{\imath} \times \hat{\jmath} = \hat{k}, \quad \hat{\jmath} \times \hat{k} = \hat{\imath}, \quad \hat{k} \times \hat{\imath} = \hat{\jmath}}^×^=k^,^×k^=^,k^×^=^ The order ı^→ȷ^→k^→ı^\hat{\imath} \to \hat{\jmath} \to \hat{k} \to \hat{\imath}^→^→k^→^ is cyclic; in reverse order the sign flips.
[Vector products] What does it mean that the cross product is anti-symmetric?
b⃗×a⃗=− a⃗×b⃗\boldsymbol{\vec{b} \times \vec{a} = -\,\vec{a} \times \vec{b}}b×a=−a×b : swapping the inputs flips the sign. So the cross product is not commutative, unlike the product of numbers ( ab=baab = baab=ba ) and the dot product ( u⃗⋅v⃗=v⃗⋅u⃗\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}u⋅v=v⋅u ).Example: ȷ^×ı^=−k^\hat{\jmath} \times \hat{\imath} = -\hat{k}^×^=−k^ , k^×ȷ^=−ı^\ \hat{k} \times \hat{\jmath} = -\hat{\imath} k^×^=−^ , ı^×k^=−ȷ^\ \hat{\imath} \times \hat{k} = -\hat{\jmath…
[Vector products] What is the formula for the cross product of a⃗=(ax,ay,az)\vec{a} = (a_x, a_y, a_z)a=(ax,ay,az) and b⃗=(bx,by,bz)\vec{b} = (b_x, b_y, b_z)b=(bx,by,bz) ?
a⃗×b⃗=(aybz−azby, azbx−axbz, axby−aybx)\boldsymbol{\vec{a} \times \vec{b} = (a_y b_z - a_z b_y, \ \ a_z b_x - a_x b_z, \ \ a_x b_y - a_y b_x)}a×b=(aybz−azby, azbx−axbz, axby−aybx) Example: (1,2,3)×(4,5,6)=(2⋅6−3⋅5, 3⋅4−1⋅6, 1⋅5−2⋅4)=(−3,6,−3)(1, 2, 3) \times (4, 5, 6) = (2 \cdot 6 - 3 \cdot 5, \ 3 \cdot 4 - 1 \cdot 6, \ 1 \cdot 5 - 2 \cdot 4) = (-3, 6, -3)(1,2,3)×(4,5,6)=(2⋅6−3⋅5, 3⋅4−1⋅6, 1⋅5−2⋅4)=(−3,6,−3) .
[Vector products] What is the length ∥a⃗×b⃗∥\|\vec{a} \times \vec{b}\|∥a×b∥ of a cross product, with φ\varphiφ the angle between a⃗\vec{a}a and b⃗\vec{b}b ?
∥a⃗×b⃗∥=∥a⃗∥∥b⃗∥sinφ\boldsymbol{\|\vec{a} \times \vec{b}\| = \|\vec{a}\| \|\vec{b}\| \sin\varphi}∥a×b∥=∥a∥∥b∥sinφ So parallel vectors have cross product 0⃗\vec{0}0 .Example: ∥ı^×ȷ^∥=1⋅1⋅sin90∘=1=∥k^∥\|\hat{\imath} \times \hat{\jmath}\| = 1 \cdot 1 \cdot \sin 90^\circ = 1 = \|\hat{k}\|∥^×^∥=1⋅1⋅sin90∘=1=∥k^∥ .
[Vector products] What does the right-hand rule say about a⃗×b⃗\vec{a} \times \vec{b}a×b ?
Index finger along a⃗\vec{a}a , middle finger along b⃗\vec{b}b : the thumb points along a⃗×b⃗\vec{a} \times \vec{b}a×b . Two vectors are perpendicular to the plane of a⃗\vec{a}a and b⃗\vec{b}b (one on each side); the right-hand rule picks the one that is a⃗×b⃗\vec{a} \times \vec{b}a×b . It agrees with ı^×ȷ^=k^\hat{\imath} \times \hat{\jmath} = \hat{k}^×^=k^ .
[Complex numbers] What are the solutions of x2+1=0x^2 + 1 = 0x2+1=0 ?
x=ix = ix=i and x=−ix = -ix=−i , where iii is the unit imaginary number, i2=−1i^2 = -1i2=−1 . There is no real solution: the graph of f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1 does not cross the xxx -axis.Check: i2+1=−1+1=0i^2 + 1 = -1 + 1 = 0i2+1=−1+1=0 and (−i)2+1=(−1)2i2+1=0(-i)^2 + 1 = (-1)^2 i^2 + 1 = 0(−i)2+1=(−1)2i2+1=0 .
[Complex numbers] What is a complex number, and what are its real part and imaginary part?
z=a+biz = a + biz=a+bi with a,b∈Ra, b \in \mathbb{R}a,b∈R ; Re{z}=a\operatorname{Re}\{z\} = aRe{z}=a , Im{z}=b\operatorname{Im}\{z\} = bIm{z}=b . C={a+bi∣a,b∈R}\mathbb{C} = \{a + bi \mid a, b \in \mathbb{R}\}C={a+bi∣a,b∈R} . A number bibibi is called imaginary. The imaginary part is the real number bbb , not bibibi .Example: z=3−4iz = 3 - 4iz=3−4i : Re{z}=3\operatorname{Re}\{z\} = 3Re{z}=3 , Im{z}=−4\operatorname{Im}\{z\} = -4Im{z}=−4 .
[Complex numbers] What is the complex conjugate zˉ\bar{z}zˉ of z=a+biz = a + biz=a+bi ?
zˉ=a−bi\boldsymbol{\bar{z} = a - bi}zˉ=a−bi Example: 3−4i‾=3+4i\overline{3 - 4i} = 3 + 4i3−4i=3+4i .
[Complex numbers] What is the complex plane, and where is z=a+biz = a + biz=a+bi in it?
A plane whose horizontal axis measures the real part and vertical axis the imaginary part; zzz is the point (vector) (a,b)(a, b)(a,b) . So complex numbers are vectors in the complex plane, with a length ∣z∣|z|∣z∣ and an angle φz\varphi_zφz .
[Complex numbers] What is the magnitude ∣z∣|z|∣z∣ of z=a+biz = a + biz=a+bi ?
∣z∣=a2+b2=zˉz\boldsymbol{|z| = \sqrt{a^2 + b^2} = \sqrt{\bar{z} z}}∣z∣=a2+b2=zˉz It is the length of the vector (a,b)(a, b)(a,b) in the complex plane (Pythagoras' theorem), and zˉz=(a−bi)(a+bi)=a2+b2\bar{z} z = (a - bi)(a + bi) = a^2 + b^2zˉz=(a−bi)(a+bi)=a2+b2 .Example: ∣3+4i∣=9+16=5|3 + 4i| = \sqrt{9 + 16} = 5∣3+4i∣=9+16=5 .
[Complex numbers] What is the phase (or argument) φz=argz\varphi_z = \arg zφz=argz of z=a+biz = a + biz=a+bi , and when is tan−1(b/a)\tan^{-1}(b/a)tan−1(b/a) correct for it?
The angle zzz makes with the real axis; tan−1(b/a)\tan^{-1}(b/a)tan−1(b/a) is correct only when a>0a > 0a>0 . tan−1\tan^{-1}tan−1 returns values in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}][−2π,2π] , so it must be corrected when a<0a < 0a<0 . The two-input function atan2(b,a)\operatorname{atan2}(b, a)atan2(b,a) gives the correct angle in all four quadrants.Example: z=−1+iz = -1 + iz=−1+i : tan−1(−1)=−π4\tan^{-1}(-1) = -\frac{\pi}{4}tan−1(−1)=−4π , but argz=atan2(1,−1)=3π4\arg z = \operat…
[Complex numbers] What is the polar representation z=∣z∣∠φzz = |z| \angle \varphi_zz=∣z∣∠φz of a complex number in terms of aaa and bbb ?
z=∣z∣∠φz=∣z∣cosφz+i ∣z∣sinφz\boldsymbol{z = |z| \angle \varphi_z = |z| \cos\varphi_z + i \, |z| \sin\varphi_z}z=∣z∣∠φz=∣z∣cosφz+i∣z∣sinφz , so a=∣z∣cosφz\boldsymbol{a = |z| \cos\varphi_z}a=∣z∣cosφz and b=∣z∣sinφz\boldsymbol{b = |z| \sin\varphi_z}b=∣z∣sinφz . Example: 2∠π2=2cosπ2+2isinπ2=2i2 \angle \frac{\pi}{2} = 2 \cos\frac{\pi}{2} + 2i \sin\frac{\pi}{2} = 2i2∠2π=2cos2π+2isin2π=2i .
[Complex numbers] How are complex numbers added?
(a+bi)+(c+di)=(a+c)+(b+d)i\boldsymbol{(a + bi) + (c + di) = (a + c) + (b + d)i}(a+bi)+(c+di)=(a+c)+(b+d)i Real parts together and imaginary parts together, like adding vectors componentwise.Example: (1+2i)+(3−5i)=4−3i(1 + 2i) + (3 - 5i) = 4 - 3i(1+2i)+(3−5i)=4−3i .
[Complex numbers] Which operations can be done with complex numbers but are not defined for vectors in R2\mathbb{R}^2R2 ?
Multiplication and division. C\mathbb{C}C is a field: addition, subtraction, multiplication and division are all defined.
[Complex numbers] How are two complex numbers multiplied in the form a+bia + bia+bi ?
(a+bi)(c+di)=(ac−bd)+(ad+bc)i\boldsymbol{(a + bi)(c + di) = (ac - bd) + (ad + bc)i}(a+bi)(c+di)=(ac−bd)+(ad+bc)i Expand with the usual rules of algebra and use i2=−1i^2 = -1i2=−1 .Example: (1+2i)(3+i)=3+i+6i+2i2=1+7i(1 + 2i)(3 + i) = 3 + i + 6i + 2i^2 = 1 + 7i(1+2i)(3+i)=3+i+6i+2i2=1+7i .
[Complex numbers] How are two complex numbers multiplied in polar representation p∠ϕp \angle \phip∠ϕ and q∠ψq \angle \psiq∠ψ ?
(p∠ϕ)(q∠ψ)=pq∠(ϕ+ψ)\boldsymbol{(p \angle \phi)(q \angle \psi) = pq \angle (\phi + \psi)}(p∠ϕ)(q∠ψ)=pq∠(ϕ+ψ) : multiply the magnitudes, add the phases. Example: i⋅(−1)=(1∠π2)(1∠π)=1∠3π2=−ii \cdot (-1) = (1 \angle \frac{\pi}{2})(1 \angle \pi) = 1 \angle \frac{3\pi}{2} = -ii⋅(−1)=(1∠2π)(1∠π)=1∠23π=−i .
[Complex numbers] What does multiplying a complex number by iii do to it in the complex plane?
It rotates it by π2\frac{\pi}{2}2π ( 90∘90^\circ90∘ ) counterclockwise. i=1∠π2i = 1 \angle \frac{\pi}{2}i=1∠2π , and multiplying multiplies magnitudes (by 1) and adds phases (adds π2\frac{\pi}{2}2π ).Example: i⋅1=ii \cdot 1 = ii⋅1=i , i⋅i=−1\ i \cdot i = -1 i⋅i=−1 , i⋅(−1)=−i\ i \cdot (-1) = -i i⋅(−1)=−i .
[Complex numbers] How do you divide a complex number zzz by a complex number s≠0s \neq 0s=0 ?
zs=z sˉ∣s∣2\boldsymbol{\dfrac{z}{s} = z \, \dfrac{\bar{s}}{|s|^2}}sz=z∣s∣2sˉ Multiply numerator and denominator by sˉ\bar{s}sˉ ; then the denominator ssˉ=∣s∣2s \bar{s} = |s|^2ssˉ=∣s∣2 is real. So sˉ∣s∣2\frac{\bar{s}}{|s|^2}∣s∣2sˉ is s−1s^{-1}s−1 .Example: 1+2i1+i=(1+2i)(1−i)2=3+i2\frac{1 + 2i}{1 + i} = \frac{(1 + 2i)(1 - i)}{2} = \frac{3 + i}{2}1+i1+2i=2(1+2i)(1−i)=23+i .
[Complex numbers] Find two numbers x1,x2x_1, x_2x1,x2 with x1+x2=10x_1 + x_2 = 10x1+x2=10 and x1x2=40x_1 x_2 = 40x1x2=40 .
x1=5+15 i,x2=5−15 i\boldsymbol{x_1 = 5 + \sqrt{15}\, i, \quad x_2 = 5 - \sqrt{15}\, i}x1=5+15i,x2=5−15i Sum: 10. Product: 25−(15)2i2=25+15=4025 - (\sqrt{15})^2 i^2 = 25 + 15 = 4025−(15)2i2=25+15=40 . No real numbers work.Gerolamo Cardano gave this example in his book Ars Magna (1545), one of the earliest uses of complex numbers.
[Complex numbers] What is the polar representation r∠φr \angle \varphir∠φ of z=−3−iz = -3 - iz=−3−i ?
z=10∠(π+0.322)\boldsymbol{z = \sqrt{10} \angle (\pi + 0.322)}z=10∠(π+0.322) r=32+12=10r = \sqrt{3^2 + 1^2} = \sqrt{10}r=32+12=10 ; tan−1(13)≈0.322\ \tan^{-1}(\frac{1}{3}) \approx 0.322 tan−1(31)≈0.322 , plus π\piπ because the real part is negative.
[Complex numbers] z=−3−i=10∠(π+0.322)z = -3 - i = \sqrt{10} \angle (\pi + 0.322)z=−3−i=10∠(π+0.322) Compute z6z^6z6 using the polar representation.
z6=1000∠1.932≈−352+936i\boldsymbol{z^6 = 1000 \angle 1.932 \approx -352 + 936i}z6=1000∠1.932≈−352+936i z6=(10)6∠6(0.322+π)=103∠(1.932+6π)z^6 = (\sqrt{10})^6 \angle 6(0.322 + \pi) = 10^3 \angle (1.932 + 6\pi)z6=(10)6∠6(0.322+π)=103∠(1.932+6π) , and multiples of 2π2\pi2π in the phase can be dropped.Check: (−3−i)2=8+6i(-3 - i)^2 = 8 + 6i(−3−i)2=8+6i , (8+6i)3=−352+936i(8 + 6i)^3 = -352 + 936i(8+6i)3=−352+936i .
[Complex numbers] What does the fundamental theorem of algebra say about a polynomial P(x)P(x)P(x) of degree nnn ?
P(x)=(x−z1)(x−z2)⋯(x−zn)P(x) = (x - z_1)(x - z_2) \cdots (x - z_n)P(x)=(x−z1)(x−z2)⋯(x−zn) for some z1,…,zn∈Cz_1, \dots, z_n \in \mathbb{C}z1,…,zn∈C , its complex roots. So every polynomial equation a0+a1x+⋯+anxn=0a_0 + a_1 x + \dots + a_n x^n = 0a0+a1x+⋯+anxn=0 has its solutions in C\mathbb{C}C .Example: x2+1=(x−i)(x+i)x^2 + 1 = (x - i)(x + i)x2+1=(x−i)(x+i) .
[Complex numbers] For each of the number sets N\mathbb{N}N , Z\mathbb{Z}Z , Q\mathbb{Q}Q , R\mathbb{R}R , C\mathbb{C}C , which kind of equation needs that set for its solutions?
N\mathbb{N}N m+n=xm + n = xm+n=x with m,n∈Nm, n \in \mathbb{N}m,n∈N Example: 2+3=x2 + 3 = x2+3=x Z\mathbb{Z}Z x+m=nx + m = nx+m=n with m,n∈Nm, n \in \mathbb{N}m,n∈N Example: x+5=2x + 5 = 2x+5=2 Q\mathbb{Q}Q mx=nm x = nmx=n with m,n∈Zm, n \in \mathbb{Z}m,n∈Z Example: 3x=23x = 23x=2 R\mathbb{R}R Example: x2=2x^2 = 2x2=2 C\mathbb{C}C Every polynomial equation Example: x2=−1x^2 = -1x2=−1 By the fundamental theorem of algebra, the chain stops at C\mathbb{C}C : no larger set of numbers is needed to s…
[Complex numbers] What is Euler's formula?
eiθ=cosθ+isinθ\boldsymbol{e^{i\theta} = \cos\theta + i \sin\theta}eiθ=cosθ+isinθ It gives another notation for the polar representation: z=∣z∣∠φz=∣z∣eiφzz = |z| \angle \varphi_z = |z| e^{i \varphi_z}z=∣z∣∠φz=∣z∣eiφz .Example: eiπ/2=cosπ2+isinπ2=ie^{i\pi/2} = \cos\frac{\pi}{2} + i \sin\frac{\pi}{2} = ieiπ/2=cos2π+isin2π=i .
[Complex numbers] What is eiπe^{i\pi}eiπ , and which equation relates eee , π\piπ , iii , 1 and 0?
eiπ=−1\boldsymbol{e^{i\pi} = -1}eiπ=−1 , so eiπ+1=0e^{i\pi} + 1 = 0eiπ+1=0 . By Euler's formula: eiπ=cosπ+isinπ=−1+0ie^{i\pi} = \cos\pi + i \sin\pi = -1 + 0ieiπ=cosπ+isinπ=−1+0i .
[Complex numbers] What does de Moivre's theorem say?
(cosθ+isinθ)n=cosnθ+isinnθ\boldsymbol{(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta}(cosθ+isinθ)n=cosnθ+isinnθ By Euler's formula, (cosθ+isinθ)n=(eiθ)n=einθ=cosnθ+isinnθ(\cos\theta + i \sin\theta)^n = (e^{i\theta})^n = e^{in\theta} = \cos n\theta + i \sin n\theta(cosθ+isinθ)n=(eiθ)n=einθ=cosnθ+isinnθ .Example: n=4n = 4n=4 , θ=π2\theta = \frac{\pi}{2}θ=2π : i4=cos2π+isin2π=1i^4 = \cos 2\pi + i \sin 2\pi = 1i4=cos2π+isin2π=1 .
[Complex numbers] How does de Moivre's theorem with n=2n = 2n=2 give the double angle formulas?
Expand (cosθ+isinθ)2(\cos\theta + i \sin\theta)^2(cosθ+isinθ)2 and compare with cos2θ+isin2θ\cos 2\theta + i \sin 2\thetacos2θ+isin2θ : cos2θ=cos2θ−sin2θ\boldsymbol{\cos 2\theta = \cos^2\theta - \sin^2\theta}cos2θ=cos2θ−sin2θ , sin2θ=2sinθcosθ\boldsymbol{\sin 2\theta = 2 \sin\theta \cos\theta}sin2θ=2sinθcosθ . (cosθ+isinθ)2=(cos2θ−sin2θ)+(2sinθcosθ)i(\cos\theta + i \sin\theta)^2 = (\cos^2\theta - \sin^2\theta) + (2 \sin\theta \cos\theta) i(cosθ+isinθ)2=(cos2θ−sin2θ)+(2sinθcosθ)i…