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 July 29th, 2019, 04:12 PM #1 Member     Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 Why I always choose vector methods Two directed line segments are given in space, direction numbers are a given, find a vector perpendicular to both. Yes, this can be done using the cross product; that's easy. But this can't be done that way. The only knowledge I have are the direction numbers associated with the given line segments (components of each vector obviously, but vector methods aren't allowed) I've tried everything. Test me, give me clues please.
 July 30th, 2019, 01:02 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Given vectors $V_i=(x_i,y_i,z_i)$, where $i=1,2$ are the known vectors. You want a vector $V_3$ so that $V1\cdot V_3=V_2\cdot V_3=0$ This is two equations in three unkowns (components of $V_3$). However, the solution is is unique up to a scalar multiplication. Thanks from topsquark
 July 30th, 2019, 01:14 PM #3 Member     Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 That's the first thing I tried. I've got my two equations in 3 unknowns but I don't know how the solution can be unique up to a scalar because it's a rectangular matrix. I made some bone idle reasoning to come up with a third row such that if it's normal to both it's normal to their sum because that then just gave me a zero row for obvious reasons of dependence. I'll read your post a few times carefully and have another go.
 July 30th, 2019, 01:26 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 The first two equations make ensure perpendicularity. The third equation basically determines the size and direction (+ or -) of the vector. V3⋅V3 = 1 is one option, though this makes the problem non-linear. You could try something like $\displaystyle x_3+y_3+z_3=1$, which is linear but not guaranteed to have a solution.
 July 31st, 2019, 01:42 PM #5 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Using $L=V_3\cdot V_3=1$ will work. Eliminate one of the components by combining the two linear equations and get a linear relation between the other two substitute into $L=1$ and get a quadratic equation in one variable. Solve and work back to get the other components.
 August 2nd, 2019, 05:16 AM #6 Member     Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 I've solved the system, but it's not giving me the correct answer as given in the book. I think the quadratic is giving me extraneous solutions.
August 2nd, 2019, 12:43 PM   #7
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Quote:
 Originally Posted by NineDivines I've solved the system, but it's not giving me the correct answer as given in the book. I think the quadratic is giving me extraneous solutions.
It would help if you showed what you got and what the book answer is.

 August 3rd, 2019, 04:21 AM #8 Member     Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 How do you enclose Latex commands here? I'm using $and$ and it's not working.
 August 3rd, 2019, 08:15 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Use $...$ for stand-alone tex, \$...\$ for inline tex.
 August 3rd, 2019, 01:27 PM #10 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 My approach: Let $V_j=(vj_1,vj_2,vj_3)$. Equations $V_k\cdot V_3=vk_1v3_1+vk_2v3_2+vk_3v3_3=0$ for $k=1,2$. Eliminate the second and third components (separately) to get $(v2_3v1_1-v1_3v2_1)v3_1+(v2_3v1_2-v1_3v2_2)v3_2=0$ and $(v2_2v1_1-v1_2v2_1)v3_1+(v2_2v1_3-v1_2v2_3)v3_3=0$ Note that the coefficients for $v3_2$ and $v3_3$ are equal of opposite sign. Let $v3_1=v2_3v1_2-v1_3v2_2$ (see note) This immediately gets $v3_2=v1_3v2_1-v2_3v1_1$ and $v3_3=v2_2v1_1-v1_2v2_1$. This (up to possible direction change) is the same as the cross product. note: Because the length of $V_3$ is free, choice of $v3_1$ is free. Thanks from NineDivines

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