Linear Algebra

§4 Vectors and Geometry

1. The Big Idea

Vectors are where algebra and geometry start speaking the same language.

A vector can be understood as:

  • an arrow with length and direction
  • a coordinate list, like (3,1,4)(3,-1,4)
  • a displacement from one point to another
  • a tool for describing lines, planes, angles, areas, volumes, and distances

The main habit is this: translate geometric words into vector operations.

Geometry wordVector idea
lengthnorm
same directionscalar multiple
perpendiculardot product is zero
normal vectorvector perpendicular to a line or plane
areacross product
volumescalar triple product
shortest distanceperpendicular component

2. Vector Basics

A vector is an object with magnitude and direction.

You can picture it as an arrow. The tail is the starting point, and the tip is the ending point.

If a vector starts at point AA and ends at point BB, we write

AB\vec{AB}

Coordinate Form

In R2\mathbb R^2, a vector has two components:

u=(u1,u2)\vec{u}=(u_1,u_2)

In R3\mathbb R^3, a vector has three components:

u=(u1,u2,u3)\vec{u}=(u_1,u_2,u_3)

In Rn\mathbb R^n, a vector has nn components:

u=(u1,u2,,un)\vec{u}=(u_1,u_2,\dots,u_n)

Vector from One Point to Another

To find the vector from AA to BB, subtract coordinates in this order:

AB=BA\vec{AB}=B-A

In R3\mathbb R^3, if

A=(x1,y1,z1)A=(x_1,y_1,z_1)

and

B=(x2,y2,z2)B=(x_2,y_2,z_2)

then

AB=(x2x1, y2y1, z2z1)\vec{AB}=(x_2-x_1,\ y_2-y_1,\ z_2-z_1)

The shortcut phrase is: terminal minus initial.

Example

If

A=(1,4,2),B=(5,1,3)A=(1,4,-2),\qquad B=(5,1,3)

then

AB=(51, 14, 3(2))=(4,3,5)\vec{AB}=(5-1,\ 1-4,\ 3-(-2))=(4,-3,5)

3. Vector Addition and Scalar Multiplication

Vector addition and subtraction happen component-by-component.

If

u=(u1,u2,,un)\vec{u}=(u_1,u_2,\dots,u_n)

and

v=(v1,v2,,vn)\vec{v}=(v_1,v_2,\dots,v_n)

then

u+v=(u1+v1,u2+v2,,un+vn)\vec{u}+\vec{v}=(u_1+v_1,u_2+v_2,\dots,u_n+v_n)

and

uv=(u1v1,u2v2,,unvn)\vec{u}-\vec{v}=(u_1-v_1,u_2-v_2,\dots,u_n-v_n)

For a scalar kk,

ku=(ku1,ku2,,kun)k\vec{u}=(ku_1,ku_2,\dots,ku_n)

Geometrically, adding vectors combines displacements. Multiplying by a scalar stretches, shrinks, or reverses a vector.


4. Length, Unit Vectors, and Parallel Vectors

Norm

The norm of a vector is its length.

For

u=(u1,u2,,un)\vec{u}=(u_1,u_2,\dots,u_n)

the norm is

u=u12+u22++un2|\vec{u}|=\sqrt{u_1^2+u_2^2+\cdots+u_n^2}

Example:

u=(3,4)\vec{u}=(3,4)

Then

u=32+42=5|\vec{u}|=\sqrt{3^2+4^2}=5

Unit Vectors

A unit vector has length 11.

To make a unit vector in the same direction as u\vec{u}, divide by the length of u\vec{u}:

uu\frac{\vec{u}}{|\vec{u}|}

This keeps the direction but removes the length.

If you want the opposite direction, use

uu-\frac{\vec{u}}{|\vec{u}|}

Parallel Vectors

Two nonzero vectors u\vec{u} and v\vec{v} are parallel if one is a scalar multiple of the other:

u=kv\vec{u}=k\vec{v}

for some scalar kk.

Scalar kkMeaning
k>0k>0same direction
k<0k<0opposite direction
k=0k=0gives the zero vector, so it is not used to compare two nonzero directions

5. Dot Product: Angles and Perpendicularity

The dot product combines two vectors and gives a number.

For

u=(u1,u2,,un)\vec{u}=(u_1,u_2,\dots,u_n)

and

v=(v1,v2,,vn)\vec{v}=(v_1,v_2,\dots,v_n)

the dot product is

uv=u1v1+u2v2++unvn\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2+\cdots+u_nv_n

The dot product is useful because it also knows about angles:

uv=uvcosθ\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta

where θ\theta is the angle between the vectors.

So

cosθ=uvuv\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}

Reading the Sign of a Dot Product

Dot productAngle typeMeaning
uv>0\vec{u}\cdot\vec{v}>0acutevectors point mostly the same way
uv=0\vec{u}\cdot\vec{v}=0right anglevectors are perpendicular
uv<0\vec{u}\cdot\vec{v}<0obtusevectors point mostly opposite ways

Orthogonal Vectors

Two vectors are orthogonal if they meet at a right angle.

Algebraically:

uv=0\vec{u}\cdot\vec{v}=0

So dot product zero means perpendicular.


6. Distance in Rn\mathbb R^n

Distance between two points is the length of the displacement vector between them.

In vector language:

d(A,B)=ABd(A,B)=|\vec{AB}|

If

A=(a1,a2,,an)A=(a_1,a_2,\dots,a_n)

and

B=(b1,b2,,bn)B=(b_1,b_2,\dots,b_n)

then

d(A,B)=(b1a1)2+(b2a2)2++(bnan)2d(A,B)=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+\cdots+(b_n-a_n)^2}

This is just the norm formula applied to AB\vec{AB}.


7. Projection

Projection means the “shadow” of one vector in the direction of another vector.

The projection of u\vec{u} onto a\vec{a} is

projau=uaa2a\operatorname{proj}_{\vec{a}}\vec{u} = \frac{\vec{u}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}

This gives the component of u\vec{u} that points in the direction of a\vec{a}.

The perpendicular component is what is left over:

uprojau\vec{u}-\operatorname{proj}_{\vec{a}}\vec{u}

Projection is useful for:

  • splitting a vector into parallel and perpendicular parts
  • closest point problems
  • distance problems
  • understanding shadows/components in a chosen direction

8. Lines and Planes: Main Ingredients

Lines and planes are built from points and direction information.

ObjectWhat you need
linea point and a direction vector
plane in point-normal forma point and a normal vector
plane in parametric forma point and two nonparallel direction vectors

A direction vector lies along the object.

A normal vector is perpendicular to the object.

This distinction is the key to most geometry problems.


9. Lines in R2\mathbb R^2: Point-Normal Form

A line in R2\mathbb R^2 can be described using a point and a normal vector.

If the line passes through

A(x0,y0)A(x_0,y_0)

and has normal vector

n=(a,b)\vec{n}=(a,b)

then the equation is

a(xx0)+b(yy0)=0a(x-x_0)+b(y-y_0)=0

This can be rewritten as

ax+by+c=0ax+by+c=0

The normal vector n=(a,b)\vec{n}=(a,b) is perpendicular to the line. That is why it appears in the equation.


10. Planes in R3\mathbb R^3: Point-Normal Form

A plane in R3\mathbb R^3 can also be described using a point and a normal vector.

If the plane passes through

A(x0,y0,z0)A(x_0,y_0,z_0)

and has normal vector

n=(a,b,c)\vec{n}=(a,b,c)

then its equation is

a(xx0)+b(yy0)+c(zz0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0

This can be rewritten as

ax+by+cz+d=0ax+by+cz+d=0

The normal vector controls the orientation of the plane. If you know a point on the plane and a vector perpendicular to the plane, you know the plane.


11. Parametric Lines and Planes

Parametric equations use one or more parameters as sliders.

Line in R3\mathbb R^3

A line needs:

  1. a point
  2. a direction vector

If a line passes through

A(x0,y0,z0)A(x_0,y_0,z_0)

and is parallel to

v=(a,b,c)\vec{v}=(a,b,c)

then

(x,y,z)=(x0,y0,z0)+t(a,b,c)(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)

So

x=x0+atx=x_0+at y=y0+bty=y_0+bt z=z0+ctz=z_0+ct

The parameter tt moves you along the line.

Symmetric Equation of a Line

If a,b,c0a,b,c\neq 0, then

xx0a=yy0b=zz0c\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}

This is the symmetric form.

Parametric Plane

A plane in parametric form needs:

  1. a point
  2. two nonparallel direction vectors

If the plane passes through

A(x0,y0,z0)A(x_0,y_0,z_0)

and is parallel to

v1=(a1,b1,c1)\vec{v}_1=(a_1,b_1,c_1)

and

v2=(a2,b2,c2)\vec{v}_2=(a_2,b_2,c_2)

then

(x,y,z)=(x0,y0,z0)+sv1+tv2(x,y,z) = (x_0,y_0,z_0) + s\vec{v}_1 + t\vec{v}_2

In components:

x=x0+a1s+a2tx=x_0+a_1s+a_2t y=y0+b1s+b2ty=y_0+b_1s+b_2t z=z0+c1s+c2tz=z_0+c_1s+c_2t

The parameters ss and tt let you move in two independent directions across the plane.


12. Cross Product

The cross product is defined only in R3\mathbb R^3.

For

u=(u1,u2,u3)\vec{u}=(u_1,u_2,u_3)

and

v=(v1,v2,v3)\vec{v}=(v_1,v_2,v_3)

the cross product is

u×v=(u2v3u3v2, u3v1u1v3, u1v2u2v1)\vec{u}\times\vec{v} = (u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1)

The output is a vector, not a number.

That vector is perpendicular to both u\vec{u} and v\vec{v}.

Cross Product Properties

u×u=0\vec{u}\times\vec{u}=\vec{0} u×v=(v×u)\vec{u}\times\vec{v}=-(\vec{v}\times\vec{u}) u(u×v)=0\vec{u}\cdot(\vec{u}\times\vec{v})=0 v(u×v)=0\vec{v}\cdot(\vec{u}\times\vec{v})=0

If u\vec{u} and v\vec{v} are parallel, then

u×v=0\vec{u}\times\vec{v}=\vec{0}

So the cross product is useful when you need a vector perpendicular to two given vectors.


13. Area and Volume

The cross product and scalar triple product turn geometry into formulas.

Area Using Cross Product

The area of the parallelogram determined by u\vec{u} and v\vec{v} is

u×v|\vec{u}\times\vec{v}|

The area of the triangle determined by u\vec{u} and v\vec{v} is

12u×v\frac12|\vec{u}\times\vec{v}|

If the triangle has points A,B,CA,B,C, use

AB\vec{AB}

and

AC\vec{AC}

Then

Area=12AB×AC\text{Area}=\frac12|\vec{AB}\times\vec{AC}|

Scalar Triple Product

For vectors u,v,w\vec{u},\vec{v},\vec{w} in R3\mathbb R^3, the scalar triple product is

u(v×w)\vec{u}\cdot(\vec{v}\times\vec{w})

It can be computed as the determinant

u1u2u3v1v2v3w1w2w3\begin{vmatrix} u_1&u_2&u_3\\ v_1&v_2&v_3\\ w_1&w_2&w_3 \end{vmatrix}

Volume of a Parallelepiped

The volume of the parallelepiped determined by u,v,w\vec{u},\vec{v},\vec{w} is

u(v×w)|\vec{u}\cdot(\vec{v}\times\vec{w})|

If the scalar triple product is zero, then the vectors lie in the same plane. That means they are coplanar.


14. Distance Formulas

Distance formulas usually come from the same idea: the shortest distance is perpendicular.

Distance from a Point to a Plane

For plane

ax+by+cz+d=0ax+by+cz+d=0

and point

P(x0,y0,z0)P(x_0,y_0,z_0)

the distance from PP to the plane is

d=ax0+by0+cz0+da2+b2+c2d= \frac{|ax_0+by_0+cz_0+d|} {\sqrt{a^2+b^2+c^2}}

The numerator measures how far the point is from satisfying the plane equation. The denominator normalizes by the length of the normal vector.

Distance from a Point to a Line in R3\mathbb R^3

Suppose a line passes through point AA and has direction vector v\vec{v}.

Let PP be another point. Then

d=v×APvd= \frac{|\vec{v}\times\vec{AP}|}{|\vec{v}|}

Why? The cross product gives the area of the parallelogram formed by v\vec{v} and AP\vec{AP}.

That area is also

base×height\text{base}\times\text{height}

The base is v|\vec{v}|, and the height is the distance.

Distance Between Skew Lines

Suppose line L1L_1 passes through A1A_1 with direction v1\vec{v}_1, and line L2L_2 passes through A2A_2 with direction v2\vec{v}_2.

First create a vector perpendicular to both lines:

n=v1×v2\vec{n}=\vec{v}_1\times\vec{v}_2

Then the distance between the skew lines is

d=A1A2nnd= \frac{|\vec{A_1A_2}\cdot\vec{n}|}{|\vec{n}|}

This measures the component of the connecting vector in the direction perpendicular to both lines.


15. Relationships Between Lines and Planes

These relationship tests are common in geometry problems.

Line and Plane

Suppose a line has direction vector v\vec{v}, and a plane has normal vector n\vec{n}.

RelationshipTestReason
line is parallel to planevn=0\vec{v}\cdot\vec{n}=0line direction is perpendicular to plane normal
line is perpendicular to planev=kn\vec{v}=k\vec{n}line direction is parallel to plane normal
line intersects planesubstitute the line into the planesolve for the parameter

Two Lines in R3\mathbb R^3

Let two lines have direction vectors v1\vec{v}_1 and v2\vec{v}_2.

RelationshipHow to check
paralleldirection vectors are scalar multiples, or v1×v2=0\vec{v}_1\times\vec{v}_2=\vec{0}
intersectingsolve the parametric equations and get a common point
skewnot parallel and no common point

Skew lines exist only in 3D. They miss each other without being parallel.


16. Closest Point Problems

Closest point problems are really perpendicularity problems.

Closest Point on a Plane

To find the closest point on a plane to a point AA:

  1. Find the normal vector n\vec{n} of the plane.
  2. Create a line through AA in direction n\vec{n}.
  3. Intersect that line with the plane.
  4. The intersection point is the closest point.

Why? The shortest path from a point to a plane is perpendicular to the plane.

Closest Point on a Line

To find the closest point on a line to point AA:

  1. Use the line direction vector v\vec{v}.
  2. Create a plane through AA whose normal vector is v\vec{v}.
  3. Intersect that plane with the line.
  4. The intersection point is the closest point.

Why? The shortest segment from a point to a line is perpendicular to the line.


17. Plane Construction Recipes

Many plane problems are about finding a normal vector. Once you have the normal vector and a point, use point-normal form.

Plane Through Three Points

Given three points A,B,CA,B,C:

  1. Compute two direction vectors:
AB\vec{AB}

and

AC\vec{AC}
  1. Cross them to get a normal vector:
n=AB×AC\vec{n}=\vec{AB}\times\vec{AC}
  1. Use point-normal form:
a(xx0)+b(yy0)+c(zz0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Plane Containing a Line and a Point

If a plane contains a line and a point not on the line:

  1. Use the direction vector of the line.
  2. Use a vector from a point on the line to the extra point.
  3. Cross those two vectors to get the normal vector.
  4. Use point-normal form.

Plane Containing Two Parallel Lines

If two lines are parallel:

  1. Use their common direction vector.
  2. Use a vector connecting a point on the first line to a point on the second line.
  3. Cross those two vectors to get the plane normal.
  4. Use point-normal form.

18. Problem-Solving Routine

For vector geometry problems, use this routine:

  1. Identify the goal. Are you finding a vector, length, angle, line, plane, area, volume, or distance?
  2. Write down the given points and vectors. Keep labels clear.
  3. If you need a direction vector, subtract points. Use terminal minus initial.
  4. If you need perpendicularity, use dot product or cross product. Dot product tests perpendicularity; cross product creates perpendicular vectors in R3\mathbb R^3.
  5. If you need a plane, look for a normal vector. A normal plus a point gives point-normal form.
  6. If you need a distance, choose the right formula. Point-plane, point-line, and skew-line distances use different setups.
  7. Check the type of answer. A distance is a scalar, a direction is a vector, a line/plane is an equation, and an intersection is a point.

19. Key Takeaways

  • A vector has magnitude and direction.
  • To find AB\vec{AB}, compute terminal minus initial.
  • The norm gives vector length.
  • A unit vector keeps direction but has length 11.
  • Dot product gives angle information and detects perpendicularity.
  • Projection gives the shadow of one vector in another direction.
  • A line needs a point and direction vector.
  • A plane in point-normal form needs a point and normal vector.
  • Cross product creates a vector perpendicular to two vectors in R3\mathbb R^3.
  • Cross product gives area; scalar triple product gives volume.
  • Shortest distance problems usually involve perpendicular directions.

Mini-Self-Check

  1. If A=(2,1,4)A=(2,-1,4) and B=(5,3,0)B=(5,3,0), what is AB\vec{AB}?

Use terminal minus initial:

AB=(52, 3(1), 04)=(3,4,4)\vec{AB}=(5-2,\ 3-(-1),\ 0-4)=(3,4,-4)
Click to reveal
  1. What does uv=0\vec{u}\cdot\vec{v}=0 tell you?

It tells you that the vectors are orthogonal, meaning they are perpendicular.

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  1. What does u×v\vec{u}\times\vec{v} give you?

It gives a vector perpendicular to both u\vec{u} and v\vec{v}, assuming the vectors are in R3\mathbb R^3.

Click to reveal
  1. What do you need to write a line in R3\mathbb R^3?

You need a point on the line and a direction vector for the line.

Click to reveal
  1. What do you need to write a plane in point-normal form?

You need a point on the plane and a normal vector perpendicular to the plane.

Click to reveal
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