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July 29th, 2019, 04:12 PM   #1
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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.
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July 30th, 2019, 01:02 PM   #2
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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.
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July 30th, 2019, 01:14 PM   #3
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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.
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July 30th, 2019, 01:26 PM   #4
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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.
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July 31st, 2019, 01:42 PM   #5
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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.
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August 2nd, 2019, 05:16 AM   #6
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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.
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August 2nd, 2019, 12:43 PM   #7
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Quote:
Originally Posted by NineDivines View Post
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.
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August 3rd, 2019, 04:21 AM   #8
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How do you enclose Latex commands here?

I'm using and it's not working.
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August 3rd, 2019, 08:15 AM   #9
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Use [math]...[/math] for stand-alone tex, \$...\$ for inline tex.
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August 3rd, 2019, 01:27 PM   #10
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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.
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