Proving Three 3D Points Are Collinear With Cross Product

Home » Collinear Vectors In 3d » Proving Three 3D Points Are Collinear With Cross Product

Maybe your like

Physics Forums Physics Forums
  • Insights Blog -- Browse All Articles -- Physics Articles Math Articles Education Articles Bio/Chem/Tech Articles
  • Forums General Math Calculus Differential Equations Topology and Analysis Linear and Abstract Algebra Differential Geometry Set Theory, Logic, Probability, Statistics MATLAB, Maple, Mathematica, LaTeX
  • Trending
Log in Register What's new
  • General Math
  • Calculus
  • Differential Equations
  • Topology and Analysis
  • Linear and Abstract Algebra
  • Differential Geometry
  • Set Theory, Logic, Probability, Statistics
  • MATLAB, Maple, Mathematica, LaTeX
Menu Log in Register Navigation More options Style variation System Light Dark Contact us Close Menu You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser.
  • Forums
  • Mathematics
  • Calculus
Proving three 3D points are collinear with Cross Product
  • Thread starter Thread starter Masschaos
  • Start date Start date May 5, 2011
  • Tags Tags 3d Cross Cross product Points Product
Click For Summary To prove that three points in 3D space are collinear, the cross product of the vectors formed by these points must equal zero. Specifically, if points A, B, and C are given, the condition c = a + t(b - a) indicates collinearity for some scalar t. The discussion clarifies that the triple scalar product a · (b x c) must also equal zero for the points to be collinear. Additionally, it is noted that using ratios of direction vectors can lead to division by zero, complicating the proof. Understanding these vector relationships is essential for confirming collinearity in three-dimensional space. Masschaos Messages 30 Reaction score 0 Hi. I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear. The best I can see is that the cross-product of the vectors must equal zero. Can someone explain how to do this a little more clearly? Physics news on Phys.org
  • New image sensor breaks optical limits
  • Ultrafast fluorescence pulse technique enables imaging of individual trapped atoms
  • Journey to the center of a quantized vortex: How microscopic mutual friction governs superfluid dissipation
Rasalhague Messages 1,383 Reaction score 2 If a, b and c are three points and there's a number t such that c = a + t(b-a), then a, b and c are collinear. Conversely, if they're collinear, then there's a number t such that c = a + t(b-a), and their triple product will equal zero: a.(b x c) = 0. You haven't defined what you mean by the cross product of three vectors. But if you mean something like "a x (b x c)", then it's not true. For example, take a = < 5, 5, 5 > b = < 1, 3, 9 > c = < -16, -8, 16 > = a + 4(b-a) a x (b x c) = < -11, -3, 21 > EDIT: I've just corrected a mistake I made. In the first version of this post, I wrote c = t(b-a), where I should have written c = a + t(b-a). Last edited: May 5, 2011 Some Pig Messages 38 Reaction score 0 (a-b)x(b-c)=0 or axb+bxc+cxa=0 Masschaos Messages 30 Reaction score 0 Ahh, I think I understand a little better. What I understand is that if I have points A, B, and C. I can find the line AB and AC. If I do the following (AB)/(AC) and find that vectors i, j and k all have the same factor. Then is that proof that they are collinear. Example AB = (1,1,1) AC = (3,3,3) (AB)/(AC) = (1/3, 1/3, 1/3). As AB and AC are all multiples of that base factor, they are collinear? Or have I gone off on a tangent? Rasalhague Messages 1,383 Reaction score 2 Yes - although careful with that method: you might get a situation where you're trying to divide by zero, and so don't get any result that way, even though the points may be collinear. (Think of what happens when the points all lie in one of the coordinate planes.) olivermsun Science Advisor Messages 1,280 Reaction score 141
Masschaos said: Example AB = (1,1,1) AC = (3,3,3) (AB)/(AC) = (1/3, 1/3, 1/3). As AB and AC are all multiples of that base factor, they are collinear?
Are you placing A is at the origin? (If you do then that's one way to make life a lot easier). LCKurtz Science Advisor Homework Helper Messages 9,567 Reaction score 775
Masschaos said: Ahh, I think I understand a little better. What I understand is that if I have points A, B, and C. I can find the line AB and AC. If I do the following (AB)/(AC) and find that vectors i, j and k all have the same factor. Then is that proof that they are collinear. Example AB = (1,1,1) AC = (3,3,3) (AB)/(AC) = (1/3, 1/3, 1/3). As AB and AC are all multiples of that base factor, they are collinear? Or have I gone off on a tangent?
I don't care for the AB/AC notation. All you need to note is AC is a constant times AB. If it wasn't, they wouldn't be collinear. Hammie Messages 111 Reaction score 0
Masschaos said: Hi. I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear. The best I can see is that the cross-product of the vectors must equal zero. Can someone explain how to do this a little more clearly?
mainly because if the points are co-linear, you have a parellpiped with volume zero.. LCKurtz Science Advisor Homework Helper Messages 9,567 Reaction score 775
Masschaos said: Hi. I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear. The best I can see is that the cross-product of the vectors must equal zero. Can someone explain how to do this a little more clearly?
Hammie said: mainly because if the points are co-linear, you have a parellpiped with volume zero..
No. For that calculation you need three coplanar vectors, and the calculation would be the triple scalar (box) product A dot (B cross C). Not the same test as for parallel vectors. Masschaos Messages 30 Reaction score 0 Ahh thanks. I understand much better now. paulfr Messages 193 Reaction score 3
Some Pig said: (a-b)x(b-c)=0 or axb+bxc+cxa=0
To be a bit more precise If a b and c are points in a plane Then <a -b> and <b-c> are direction vectors in the plane If the points are on the same line, then the angle between them is 0 degrees Since <a -b> x <b-c> = || <a -b> || || <b-c> || sin x If x = 0 then sin x = 0 so <a -b> x <b-c> = 0

Similar threads

Cross Product of Parallel Vectors is the zero vector (why?)
  • Sep 3, 2015 · Replies 11 · Sep 3, 2015
Replies 11 Views 17K Undergrad Why is the Cross Product Used in Mathematics? Understanding its Role and History
  • Dec 12, 2023 · Replies 21 · Nov 26, 2024
Replies 21 Views 3K Why Is the Cross Product Perpendicular to Its Constituent Vectors?
  • Aug 13, 2013 · Replies 26 · Mar 14, 2015
Replies 26 Views 18K Is the Cross Product Valid in Higher Dimensions?
  • Jan 17, 2013 · Replies 9 · Jan 18, 2013
Replies 9 Views 4K Graduate Reason out the cross product (for the moment): a skew symmetric form
  • Jan 4, 2022 · Replies 14 · Jan 4, 2022
Replies 14 Views 3K High School A "no hands" rule for the cross product (requires literacy)
  • Nov 4, 2021 · Replies 8 · Nov 5, 2021
Replies 8 Views 2K Undergrad Can the cross product concept be completely replaced by the exterior product?
  • Apr 11, 2019 · Replies 5 · Apr 20, 2019
Replies 5 Views 2K Cross Products Verifying Third Vector Orthoginality?
  • Mar 6, 2013 · Replies 4 · Mar 8, 2013
Replies 4 Views 3K Undergrad Cross Product in E_u: Explaining Gourgoulhon's Text
  • Feb 23, 2021 · Replies 4 · Feb 26, 2021
Replies 4 Views 2K Graduate Tensor product matrices order relation
  • May 14, 2022 · Replies 1 · May 26, 2022
Replies 1 Views 1K
  • Forums
  • Mathematics
  • Calculus

Hot Threads

  • M

    Undergrad Problem in understanding instantaneous velocity

    • Started by Mike_bb
    • Jul 11, 2025
    • Replies: 33
    • Calculus
  • aronclark1017

    Undergrad Unit Circle Confusion: A Self-Study Challenge?

    • Started by aronclark1017
    • Sep 29, 2025
    • Replies: 7
    • Calculus
  • H

    Undergrad Proving that convexity implies second order derivative being positive

    • Started by hmparticle9
    • Dec 4, 2025
    • Replies: 6
    • Calculus
  • H

    Undergrad Derive the orthonormality condition for Legendre polynomials

    • Started by hmparticle9
    • Jul 18, 2025
    • Replies: 30
    • Calculus
  • M

    Undergrad Problem with calculating projections of curl using rotation of contour

    • Started by Mike_bb
    • Nov 22, 2025
    • Replies: 9
    • Calculus

Recent Insights

  • Greg Bernhardt

    Insights Thinking Outside The Box Versus Knowing What’s In The Box

    • Started by Greg Bernhardt
    • Oct 13, 2025
    • Replies: 26
    • Other Physics Topics
  • Greg Bernhardt

    Insights Why Entangled Photon-Polarization Qubits Violate Bell’s Inequality

    • Started by Greg Bernhardt
    • Sep 29, 2025
    • Replies: 28
    • Quantum Interpretations and Foundations
  • Greg Bernhardt

    Insights Quantum Entanglement is a Kinematic Fact, not a Dynamical Effect

    • Started by Greg Bernhardt
    • Sep 2, 2025
    • Replies: 22
    • Quantum Physics
  • Greg Bernhardt

    Insights What Exactly is Dirac’s Delta Function? - Insight

    • Started by Greg Bernhardt
    • Sep 2, 2025
    • Replies: 33
    • General Math
  • Greg Bernhardt

    Insights Relativator (Circular Slide-Rule): Simulated with Desmos - Insight

    • Started by Greg Bernhardt
    • Sep 2, 2025
    • Replies: 1
    • Special and General Relativity
  • P

    Insights Fixing Things Which Can Go Wrong With Complex Numbers

    • Started by PAllen
    • Jul 20, 2025
    • Replies: 7
    • General Math
Back Top

Tag » Collinear Vectors In 3d