User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

 February 4th, 2018, 04:37 PM #1 Newbie   Joined: Feb 2018 From: Canada Posts: 1 Thanks: 0 Negative determinant is positive? Hello, I am looking at this question. "Determine a general form of the equation of the plane through each of the following sets of three points" P1(1,-2,3) P2(1,0,2) P3(-1,4,6) This seems simple enough. I get this matrix Figuring out the determinants for each will get me 12,-2,4 So putting them in point normal form SHOULD net me 12(x-1)-2(y+2)+4(z-3) And the general form of that would be 12x-12-2y-4+4z-12 Which becomes 12x-2y+4z-28 = 0 Simplified to 6x-y+2z-14 = 0 However this answer is WRONG. The actual answer is 6x+y+2z-10 = 0 The mistake I made is that the point normal form should be 12(x-1)+2(y+2)+4(z-3) But I don't understand why. The determinant of the matrix 0 -1 -2 3 Is unquestionably -2. So why is the point normal form using +2 instead? I am especially confused because I have done other questions like this and have gotten answer correct without having to switch around signs. So what is happening here that I'm not seeing? Help would be much appreciated, thank you. Last edited by skipjack; February 5th, 2018 at 12:58 AM. February 5th, 2018, 12:39 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 I admit I don't really follow what you've done. The way I would attack this is form $v_1 = p_2-p_1,~v_2 = p_3-p_1$ $v_1 = (0,2,-1),~v_2 = (-2,6,3)$ $n = v_1 \times v_2 = (12,2,4)$ $n$ is the normal vector to the plane Thus, using $p_1$ as a point on the plane, the scalar equation of the plane is given by $n\cdot ((x,y,z)-p_1) = 0$ or $12(x-1)+2(y+2) + 4(z-3)=0$ or $12x + 2y + 4z = 20$ which reduces to $6x + y + 2z = 10$ February 5th, 2018, 01:42 AM   #3
Global Moderator

Joined: Dec 2006

Posts: 20,969
Thanks: 2219

Quote:
 Originally Posted by MontanaMax So why is the point normal form using +2 instead?
You should always alternate the signs. That's how determinant evaluation is done.

For the benefit of others, your method is briefly described here. February 5th, 2018, 06:01 AM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics Your 2-by-2 determinant computation is correct as you noticed but your 3-by-3 determinant is where you are making the mistake. Namely, as you compute minors for the matrix they should alternate in sign. The second minor is -2 as you noticed but should gain a minus sign from the determinant computation of the 3-by-3 matrix. Read here for a reminder on how to compute determinants. https://en.wikipedia.org/wiki/Determinant Tags determinant, negative, positive Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post QEDboi Algebra 3 September 1st, 2017 07:20 AM Opposite Algebra 2 September 17th, 2014 09:27 PM SSJBartSimp Calculus 1 June 11th, 2012 10:56 AM xinglongdada Linear Algebra 1 May 7th, 2012 11:35 PM David_Lete Algebra 3 April 9th, 2009 05:03 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      