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The question I'm working on is:

Two planes have non-parallel unit normals n and m and their closest distances from the origin are 3 and 7 respectively. Find the vector equation of their line of intersection.

I have calculated the implicit vector given by the cross product of the two normals, and from there derived a point on the line, but this is all done implicitly and seems rather vague. #

I'm thinking there may be something in the fact that the normals given are unit vectors, but I've become rather stuck.

Any help will be greatly appreciated.

Thanks

Jbo
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1 Answers1

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Hint: if $\vec r$ is a positional vector, then the equations of the planes are $\vec n \cdot \vec r=3$ and $\vec m \cdot \vec r=7$. Does this help?

Vasili
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  • Thanks for the response Vasya. I did touch on that earlier in my rough workings but ended up down a dead end, probably because I’d already worked myself into a stress. I’ll take it up again tomorrow morning and hopefully have more luck. Thanks for the hint. – Jbo Apr 15 '21 at 21:09
  • I’ve tried to pursue the angle of letting r=(x,y,z) and n = (n1,n2,n3) similarly for m. The direction vector, q, for the line of intersection must be the cross product of n and m, and also will have dot product = 0 with n and m, then I ended up with a system of equations which didn’t tell me much. Is there something to gain from considering the form (r-a)•n=0, and if so, in this form could I also choose the arbitrary vector a to be the same vector to give (r-a)•m=0? Thanks – Jbo Apr 16 '21 at 03:47