Timedora

1 Vectors and Geometry

A progressive guide to representing vectors, performing vector operations, measuring length and angles, and using vectors to describe lines and planes in two and three dimensions.

Representing and Operating on Vectors

Vectors can be understood in two equivalent ways: geometrically as directed arrows and algebraically as ordered lists of numbers. A vector from P=(p1,p2)P=(p_1,p_2) to Q=(q1,q2)Q=(q_1,q_2) is found by subtracting the initial point from the terminal point:

PQ→=Q−P=⟨q1−p1,q2−p2⟩.\overrightarrow{PQ}=Q-P=\langle q_1-p_1,q_2-p_2\rangle.

In three dimensions, the same process gives

PQ→=⟨q1−p1,q2−p2,q3−p3⟩.\overrightarrow{PQ}=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle.

An algebraic vector in Rn\mathbb R^n is an ordered tuple v=⟨v1,v2,…,vn⟩\mathbf v=\langle v_1,v_2,\ldots,v_n\rangle. The zero vector is 0=⟨0,0,…,0⟩\mathbf0=\langle0,0,\ldots,0\rangle. A can be moved without changing its identity as long as its direction and magnitude stay the same.

Componentwise operations

For vectors u=⟨u1,u2,u3⟩\mathbf u=\langle u_1,u_2,u_3\rangle and v=⟨v1,v2,v3⟩\mathbf v=\langle v_1,v_2,v_3\rangle, add and subtract corresponding components:

u+v=⟨u1+v1,u2+v2,u3+v3⟩,\mathbf u+\mathbf v=\langle u_1+v_1,u_2+v_2,u_3+v_3\rangle,
u−v=⟨u1−v1,u2−v2,u3−v3⟩.\mathbf u-\mathbf v=\langle u_1-v_1,u_2-v_2,u_3-v_3\rangle.

Scalar multiplication also acts componentwise:

cv=⟨cv1,cv2,cv3⟩.c\mathbf v=\langle cv_1,cv_2,cv_3\rangle.

If c>0c>0, the direction is preserved; if c<0c<0, the direction is reversed; and the magnitude is multiplied by ∣c∣|c|. For example, if u=⟨2,−1,3⟩\mathbf u=\langle2,-1,3\rangle and v=⟨−1,4,2⟩\mathbf v=\langle-1,4,2\rangle, then

u+v=⟨1,3,5⟩\mathbf u+\mathbf v=\langle1,3,5\rangle

and

2u−v=⟨5,−6,4⟩.2\mathbf u-\mathbf v=\langle5,-6,4\rangle.

Takeaway: Subtracting points gives a displacement vector, and vector arithmetic is performed coordinate by coordinate.

Linear Combinations, , and Independence

A linear combination is formed by multiplying vectors by scalars and adding the results:

c1v1+c2v2+⋯+ckvk.c_1\mathbf v_1+c_2\mathbf v_2+\cdots+c_k\mathbf v_k.

For example, with u=⟨1,2⟩\mathbf u=\langle1,2\rangle and v=⟨3,−1⟩\mathbf v=\langle3,-1\rangle,

2u−v=2⟨1,2⟩−⟨3,−1⟩=⟨−1,5⟩.2\mathbf u-\mathbf v=2\langle1,2\rangle-\langle3,-1\rangle=\langle-1,5\rangle.

The is the complete set of vectors obtainable from all allowed linear combinations. A single nonzero vector spans a line through the origin. Two nonparallel vectors in R3\mathbb R^3 a plane through the origin. If one vector is a scalar multiple of another, they only a line.

Linear independence describes whether any vector in a collection can be produced from the others. The vectors v1,…,vk\mathbf v_1,\ldots,\mathbf v_k are when

c1v1+⋯+ckvk=0c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf0

forces every coefficient to be zero. Otherwise, the vectors are linearly dependent.

The number of independent directions determines the geometric object described by a linear combination:

  • one independent direction gives a line;

  • two independent directions give a plane;

  • three independent directions in R3\mathbb R^3 can generate three-dimensional space.

Takeaway: Linear combinations connect algebra to geometry: coefficients control movement along available directions, while independence determines how many genuinely different directions are present.

Length, Distance, Angles, and Orthogonality

The measures the length of a vector. For v=⟨v1,v2,…,vn⟩\mathbf v=\langle v_1,v_2,\ldots,v_n\rangle,

∥v∥=v12+v22+⋯+vn2.\|\mathbf v\|=\sqrt{v_1^2+v_2^2+\cdots+v_n^2}.

For example,

∥⟨3,−4,12⟩∥=32+(−4)2+122=13.\|\langle3,-4,12\rangle\|=\sqrt{3^2+(-4)^2+12^2}=13.

The distance between points PP and QQ is the norm of their difference:

d(P,Q)=∥Q−P∥.d(P,Q)=\|Q-P\|.

For a nonzero vector, normalization produces a in the same direction:

v^=v∥v∥.\widehat{\mathbf v}=\frac{\mathbf v}{\|\mathbf v\|}.

The combines components by multiplication and addition:

u⋅v=u1v1+u2v2+⋯+unvn.\mathbf u\cdot\mathbf v=u_1v_1+u_2v_2+\cdots+u_nv_n.

It also links components to the angle θ\theta between nonzero vectors:

u⋅v=∥u∥∥v∥cos⁡θ.\mathbf u\cdot\mathbf v=\|\mathbf u\|\|\mathbf v\|\cos\theta.

Therefore,

cos⁡θ=u⋅v∥u∥∥v∥.\cos\theta=\frac{\mathbf u\cdot\mathbf v}{\|\mathbf u\|\|\mathbf v\|}.

The is positive for an acute angle, zero for a right angle, and negative for an obtuse angle. For example, if u=⟨1,2,2⟩\mathbf u=\langle1,2,2\rangle and v=⟨2,1,−2⟩\mathbf v=\langle2,1,-2\rangle, then

u⋅v=1(2)+2(1)+2(−2)=0,\mathbf u\cdot\mathbf v=1(2)+2(1)+2(-2)=0,

so the vectors are .

The same operation models work: if a constant force F\mathbf F produces displacement d\mathbf d, then

W=F⋅d.W=\mathbf F\cdot\mathbf d.

Takeaway: The norm measures size, while the measures alignment and provides a test for perpendicularity.

Projections and Vector Components

Projection separates a vector into a part parallel to a chosen direction and a part perpendicular to that direction. For a nonzero vector u\mathbf u, the of v\mathbf v onto u\mathbf u is

proj⁡uv=v⋅u∥u∥2u.\operatorname{proj}_{\mathbf u}\mathbf v=\frac{\mathbf v\cdot\mathbf u}{\|\mathbf u\|^2}\mathbf u.

The scalar projection, or signed component, is

comp⁡uv=v⋅u∥u∥.\operatorname{comp}_{\mathbf u}\mathbf v=\frac{\mathbf v\cdot\mathbf u}{\|\mathbf u\|}.

The decomposition is

v=proj⁡uv+(v−proj⁡uv).\mathbf v=\operatorname{proj}_{\mathbf u}\mathbf v+\left(\mathbf v-\operatorname{proj}_{\mathbf u}\mathbf v\right).

For v=⟨3,4⟩\mathbf v=\langle3,4\rangle and u=⟨1,0⟩\mathbf u=\langle1,0\rangle,

v⋅u=3,∥u∥2=1,\mathbf v\cdot\mathbf u=3, \qquad \|\mathbf u\|^2=1,

so

proj⁡uv=3⟨1,0⟩=⟨3,0⟩.\operatorname{proj}_{\mathbf u}\mathbf v=3\langle1,0\rangle=\langle3,0\rangle.

The perpendicular component is

v−proj⁡uv=⟨0,4⟩.\mathbf v-\operatorname{proj}_{\mathbf u}\mathbf v=\langle0,4\rangle.

Thus, ⟨3,4⟩\langle3,4\rangle is decomposed into a component along the horizontal direction and a component perpendicular to it.

Takeaway: Projection extracts the amount of a vector that lies along a specified direction; subtracting that projection leaves an remainder.

Lines, Planes, and Normal Directions

A line is determined by one point and one nonzero direction vector. If it passes through a point with position vector r0\mathbf r_0 and has direction d≠0\mathbf d\neq\mathbf0, its vector equation is

r(t)=r0+td,t∈R.\mathbf r(t)=\mathbf r_0+t\mathbf d, \qquad t\in\mathbb R.

If P0=(x0,y0,z0)P_0=(x_0,y_0,z_0) and d=⟨a,b,c⟩\mathbf d=\langle a,b,c\rangle, then

⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨a,b,c⟩,\langle x,y,z\rangle=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle,

which gives

x=x0+at,y=y0+bt,z=z0+ct.x=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct.

For the line through P=(1,−2,3)P=(1,-2,3) and Q=(4,0,−1)Q=(4,0,-1), the direction is

PQ→=Q−P=⟨3,2,−4⟩.\overrightarrow{PQ}=Q-P=\langle3,2,-4\rangle.

Therefore,

r(t)=⟨1,−2,3⟩+t⟨3,2,−4⟩,\mathbf r(t)=\langle1,-2,3\rangle+t\langle3,2,-4\rangle,

or equivalently

x=1+3t,y=−2+2t,z=3−4t.x=1+3t,\qquad y=-2+2t,\qquad z=3-4t.

A plane is determined by a point and a . If the plane passes through P0=(x0,y0,z0)P_0=(x_0,y_0,z_0) and has n=⟨A,B,C⟩\mathbf n=\langle A,B,C\rangle, then every point P=(x,y,z)P=(x,y,z) on it satisfies

n⋅(P−P0)=0.\mathbf n\cdot(P-P_0)=0.

In coordinates, this becomes

A(x−x0)+B(y−y0)+C(z−z0)=0.A(x-x_0)+B(y-y_0)+C(z-z_0)=0.

After expansion, the equation has the form

Ax+By+Cz=D.Ax+By+Cz=D.

For example, the plane through P0=(1,2,−1)P_0=(1,2,-1) with n=⟨2,−3,4⟩\mathbf n=\langle2,-3,4\rangle is

2(x−1)−3(y−2)+4(z+1)=0,2(x-1)-3(y-2)+4(z+1)=0,

which simplifies to

2x−3y+4z+8=0.2x-3y+4z+8=0.

If three noncollinear points P,Q,RP,Q,R are given, form two in-plane vectors, u=Q−P\mathbf u=Q-P and v=R−P\mathbf v=R-P. Their n=u×v\mathbf n=\mathbf u\times\mathbf v is perpendicular to the plane, so the plane can be written as

n⋅(X−P)=0.\mathbf n\cdot(X-P)=0.

Takeaway: One direction parameterizes a line, while two independent directions parameterize a plane; a plane equation is most naturally built from a point and a perpendicular direction.