evan's notes
cal 3discreteodeenv systems

On this page

  • 1. The Big Idea
  • 2. Vector Basics
  • Coordinate Form
  • Vector from One Point to Another
  • Example
  • 3. Vector Addition and Scalar Multiplication
  • 4. Length, Unit Vectors, and Parallel Vectors
  • Norm
  • Unit Vectors
  • Parallel Vectors
  • 5. Dot Product: Angles and Perpendicularity
  • Reading the Sign of a Dot Product
  • Orthogonal Vectors
  • 6. Distance in Rn\mathbb R^nRn
  • 7. Projection
  • 8. Lines and Planes: Main Ingredients
  • 9. Lines in R2\mathbb R^2R2: Point-Normal Form
  • 10. Planes in R3\mathbb R^3R3: Point-Normal Form
  • 11. Parametric Lines and Planes
  • Line in R3\mathbb R^3R3
  • Symmetric Equation of a Line
  • Parametric Plane
  • 12. Cross Product
  • Cross Product Properties
  • 13. Area and Volume
  • Area Using Cross Product
  • Scalar Triple Product
  • Volume of a Parallelepiped
  • 14. Distance Formulas
  • Distance from a Point to a Plane
  • Distance from a Point to a Line in R3\mathbb R^3R3
  • Distance Between Skew Lines
  • 15. Relationships Between Lines and Planes
  • Line and Plane
  • Two Lines in R3\mathbb R^3R3
  • 16. Closest Point Problems
  • Closest Point on a Plane
  • Closest Point on a Line
  • 17. Plane Construction Recipes
  • Plane Through Three Points
  • Plane Containing a Line and a Point
  • Plane Containing Two Parallel Lines
  • 18. Problem-Solving Routine
  • 19. Key Takeaways
  • Mini-Self-Check

Linear Algebra

§4 Vectors and Geometry

Evan Luo · Feb 24, 2026

Linear Algebra

§4 Vectors and Geometry

Evan LuoFeb 24, 2026

14 min read

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)(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 AAA and ends at point BBB, we write

AB⃗\vec{AB} AB

Coordinate Form

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

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

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

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

In Rn\mathbb R^nRn, a vector has nnn components:

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

Vector from One Point to Another

To find the vector from AAA to BBB, subtract coordinates in this order:

AB⃗=B−A\vec{AB}=B-A AB=B−A

In R3\mathbb R^3R3, if

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

and

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

then

AB⃗=(x2−x1, y2−y1, z2−z1)\vec{AB}=(x_2-x_1,\ y_2-y_1,\ z_2-z_1) AB=(x2​−x1​, y2​−y1​, z2​−z1​)

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) A=(1,4,−2),B=(5,1,3)

then

AB⃗=(5−1, 1−4, 3−(−2))=(4,−3,5)\vec{AB}=(5-1,\ 1-4,\ 3-(-2))=(4,-3,5) 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

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

and

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

then

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

and

u⃗−v⃗=(u1−v1,u2−v2,…,un−vn)\vec{u}-\vec{v}=(u_1-v_1,u_2-v_2,\dots,u_n-v_n) u−v=(u1​−v1​,u2​−v2​,…,un​−vn​)

For a scalar kkk,

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

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

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

the norm is

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

Example:

u⃗=(3,4)\vec{u}=(3,4) u=(3,4)

Then

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

Unit Vectors

A unit vector has length 111.

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

u⃗∣u⃗∣\frac{\vec{u}}{|\vec{u}|} ∣u∣u​

This keeps the direction but removes the length.

If you want the opposite direction, use

−u⃗∣u⃗∣-\frac{\vec{u}}{|\vec{u}|} −∣u∣u​

Parallel Vectors

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

u⃗=kv⃗\vec{u}=k\vec{v} u=kv

for some scalar kkk.

Scalar kkkMeaning
k>0k>0k>0same direction
k<0k<0k<0opposite direction
k=0k=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

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

and

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

the dot product is

u⃗⋅v⃗=u1v1+u2v2+⋯+unvn\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2+\cdots+u_nv_n u⋅v=u1​v1​+u2​v2​+⋯+un​vn​

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

u⃗⋅v⃗=∣u⃗∣∣v⃗∣cos⁡θ\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta u⋅v=∣u∣∣v∣cosθ

where θ\thetaθ is the angle between the vectors.

So

cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|} cosθ=∣u∣∣v∣u⋅v​

Reading the Sign of a Dot Product

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

Orthogonal Vectors

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

Algebraically:

u⃗⋅v⃗=0\vec{u}\cdot\vec{v}=0 u⋅v=0

So dot product zero means perpendicular.


6. Distance in Rn\mathbb R^nRn

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

In vector language:

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

If

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

and

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

then

d(A,B)=(b1−a1)2+(b2−a2)2+⋯+(bn−an)2d(A,B)=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+\cdots+(b_n-a_n)^2} d(A,B)=(b1​−a1​)2+(b2​−a2​)2+⋯+(bn​−an​)2​

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


7. Projection

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

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

proj⁡a⃗u⃗=u⃗⋅a⃗∣a⃗∣2a⃗\operatorname{proj}_{\vec{a}}\vec{u} = \frac{\vec{u}\cdot\vec{a}}{|\vec{a}|^2}\vec{a} proja​u=∣a∣2u⋅a​a

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

The perpendicular component is what is left over:

u⃗−proj⁡a⃗u⃗\vec{u}-\operatorname{proj}_{\vec{a}}\vec{u} u−proja​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^2R2: Point-Normal Form

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

If the line passes through

A(x0,y0)A(x_0,y_0) A(x0​,y0​)

and has normal vector

n⃗=(a,b)\vec{n}=(a,b) n=(a,b)

then the equation is

a(x−x0)+b(y−y0)=0a(x-x_0)+b(y-y_0)=0 a(x−x0​)+b(y−y0​)=0

This can be rewritten as

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

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


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

A plane in R3\mathbb R^3R3 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) A(x0​,y0​,z0​)

and has normal vector

n⃗=(a,b,c)\vec{n}=(a,b,c) n=(a,b,c)

then its equation is

a(x−x0)+b(y−y0)+c(z−z0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0 a(x−x0​)+b(y−y0​)+c(z−z0​)=0

This can be rewritten as

ax+by+cz+d=0ax+by+cz+d=0 ax+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^3R3

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) A(x0​,y0​,z0​)

and is parallel to

v⃗=(a,b,c)\vec{v}=(a,b,c) 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) (x,y,z)=(x0​,y0​,z0​)+t(a,b,c)

So

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

The parameter ttt moves you along the line.

Symmetric Equation of a Line

If a,b,c≠0a,b,c\neq 0a,b,c=0, then

x−x0a=y−y0b=z−z0c\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c} ax−x0​​=by−y0​​=cz−z0​​

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) A(x0​,y0​,z0​)

and is parallel to

v⃗1=(a1,b1,c1)\vec{v}_1=(a_1,b_1,c_1) v1​=(a1​,b1​,c1​)

and

v⃗2=(a2,b2,c2)\vec{v}_2=(a_2,b_2,c_2) v2​=(a2​,b2​,c2​)

then

(x,y,z)=(x0,y0,z0)+sv⃗1+tv⃗2(x,y,z) = (x_0,y_0,z_0) + s\vec{v}_1 + t\vec{v}_2 (x,y,z)=(x0​,y0​,z0​)+sv1​+tv2​

In components:

x=x0+a1s+a2tx=x_0+a_1s+a_2t x=x0​+a1​s+a2​t y=y0+b1s+b2ty=y_0+b_1s+b_2t y=y0​+b1​s+b2​t z=z0+c1s+c2tz=z_0+c_1s+c_2t z=z0​+c1​s+c2​t

The parameters sss and ttt let you move in two independent directions across the plane.


12. Cross Product

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

For

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

and

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

the cross product is

u⃗×v⃗=(u2v3−u3v2, u3v1−u1v3, u1v2−u2v1)\vec{u}\times\vec{v} = (u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1) u×v=(u2​v3​−u3​v2​, u3​v1​−u1​v3​, u1​v2​−u2​v1​)

The output is a vector, not a number.

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

Cross Product Properties

u⃗×u⃗=0⃗\vec{u}\times\vec{u}=\vec{0} u×u=0 u⃗×v⃗=−(v⃗×u⃗)\vec{u}\times\vec{v}=-(\vec{v}\times\vec{u}) u×v=−(v×u) u⃗⋅(u⃗×v⃗)=0\vec{u}\cdot(\vec{u}\times\vec{v})=0 u⋅(u×v)=0 v⃗⋅(u⃗×v⃗)=0\vec{v}\cdot(\vec{u}\times\vec{v})=0 v⋅(u×v)=0

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

u⃗×v⃗=0⃗\vec{u}\times\vec{v}=\vec{0} u×v=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 u⃗\vec{u}u and v⃗\vec{v}v is

∣u⃗×v⃗∣|\vec{u}\times\vec{v}| ∣u×v∣

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

12∣u⃗×v⃗∣\frac12|\vec{u}\times\vec{v}| 21​∣u×v∣

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

AB⃗\vec{AB} AB

and

AC⃗\vec{AC} AC

Then

Area=12∣AB⃗×AC⃗∣\text{Area}=\frac12|\vec{AB}\times\vec{AC}| Area=21​∣AB×AC∣

Scalar Triple Product

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

u⃗⋅(v⃗×w⃗)\vec{u}\cdot(\vec{v}\times\vec{w}) u⋅(v×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} ​u1​v1​w1​​u2​v2​w2​​u3​v3​w3​​​

Volume of a Parallelepiped

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

∣u⃗⋅(v⃗×w⃗)∣|\vec{u}\cdot(\vec{v}\times\vec{w})| ∣u⋅(v×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 ax+by+cz+d=0

and point

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

the distance from PPP to the plane is

d=∣ax0+by0+cz0+d∣a2+b2+c2d= \frac{|ax_0+by_0+cz_0+d|} {\sqrt{a^2+b^2+c^2}} d=a2+b2+c2​∣ax0​+by0​+cz0​+d∣​

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^3R3

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

Let PPP be another point. Then

d=∣v⃗×AP⃗∣∣v⃗∣d= \frac{|\vec{v}\times\vec{AP}|}{|\vec{v}|} d=∣v∣∣v×AP∣​

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

That area is also

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

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

Distance Between Skew Lines

Suppose line L1L_1L1​ passes through A1A_1A1​ with direction v⃗1\vec{v}_1v1​, and line L2L_2L2​ passes through A2A_2A2​ with direction v⃗2\vec{v}_2v2​.

First create a vector perpendicular to both lines:

n⃗=v⃗1×v⃗2\vec{n}=\vec{v}_1\times\vec{v}_2 n=v1​×v2​

Then the distance between the skew lines is

d=∣A1A2⃗⋅n⃗∣∣n⃗∣d= \frac{|\vec{A_1A_2}\cdot\vec{n}|}{|\vec{n}|} d=∣n∣∣A1​A2​​⋅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 v⃗\vec{v}v, and a plane has normal vector n⃗\vec{n}n.

RelationshipTestReason
line is parallel to planev⃗⋅n⃗=0\vec{v}\cdot\vec{n}=0v⋅n=0line direction is perpendicular to plane normal
line is perpendicular to planev⃗=kn⃗\vec{v}=k\vec{n}v=knline direction is parallel to plane normal
line intersects planesubstitute the line into the planesolve for the parameter

Two Lines in R3\mathbb R^3R3

Let two lines have direction vectors v⃗1\vec{v}_1v1​ and v⃗2\vec{v}_2v2​.

RelationshipHow to check
paralleldirection vectors are scalar multiples, or v⃗1×v⃗2=0⃗\vec{v}_1\times\vec{v}_2=\vec{0}v1​×v2​=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 AAA:

  1. Find the normal vector n⃗\vec{n}n of the plane.
  2. Create a line through AAA in direction n⃗\vec{n}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 AAA:

  1. Use the line direction vector v⃗\vec{v}v.
  2. Create a plane through AAA whose normal vector is v⃗\vec{v}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,CA,B,C:

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

and

AC⃗\vec{AC} AC
  1. Cross them to get a normal vector:
n⃗=AB⃗×AC⃗\vec{n}=\vec{AB}\times\vec{AC} n=AB×AC
  1. Use point-normal form:
a(x−x0)+b(y−y0)+c(z−z0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0 a(x−x0​)+b(y−y0​)+c(z−z0​)=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^3R3.
  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 AB⃗\vec{AB}AB, compute terminal minus initial.
  • The norm gives vector length.
  • A unit vector keeps direction but has length 111.
  • 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^3R3.
  • 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)A=(2,−1,4) and B=(5,3,0)B=(5,3,0)B=(5,3,0), what is AB⃗\vec{AB}AB?

Use terminal minus initial:

AB⃗=(5−2, 3−(−1), 0−4)=(3,4,−4)\vec{AB}=(5-2,\ 3-(-1),\ 0-4)=(3,4,-4) AB=(5−2, 3−(−1), 0−4)=(3,4,−4)
Click to reveal
  1. What does u⃗⋅v⃗=0\vec{u}\cdot\vec{v}=0u⋅v=0 tell you?

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

Click to reveal
  1. What does u⃗×v⃗\vec{u}\times\vec{v}u×v give you?

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

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

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

Source: https://notes.ohevan.com/notes/linear/04-vectors-and-geometry

© 2026 Evan Luo. All rights reserved.

Back to Linear Algebra

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·Last edited May 23, 2026
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© 2026 Evan Luo. All rights reserved.