Where Line Meets Plane
Consider a plane passing through the point ( 1 , 2 , 3 ) with normal vector ( 4 , 5 , 6 ) . Suppose we start at the point ( − 1 , 0 , 5 ) and travel in the direction of the vector ( − 1 , − 1 , 2 ) until we intersect the plane.
If the intersection point is ( a , b , c ) , what is a + b + c ?
The answer is 4.
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1 solution
We do not have to solve for t as x+y+z will always be equal to 4.
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uh huh, so you caught on right away An upvote for your comment, one of the shortest solutions ever.
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Interesting. So I just happened to pick a vector whose components added to zero, making this problem short-cuttable.
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@Steven Chase – That's right. I mean, I could pose a problem where given a point ( a , b , c ) , going from there by some vector ( r , s , t ) (where r+s+t is secretly = 0), and ask for the sum of the coordinates of the point where it intersects with, say, a icosahedron. That'll get problem-solvers to scratching their heads, except for people like Archit Agrawal.
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@Michael Mendrin – Indeed. That's what happens when you start out thinking symbolically about the general case, decide on what your solution approach needs to be, and only then look at the actual numbers.
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@Steven Chase – So the answer was unintentionally hidden in plain sight. Given this short-cut, then, I find it amusing that this problem has achieved Level 5 status. Perhaps another similar posting with a less accommodating vector is in order? :)
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@Brian Charlesworth – Perhaps so. Although how many people do you think are going to find the shortcut :)? I'd wager not all that many. We apparently didn't.
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@Steven Chase – Probably not many, especially when they start out seeing that it's a level 5 problem and thus aren't anticipating any potential short-cuts. It took me a while to get out of the "symbolic" mindset and just look at the parametric equations to see the obvious fact that Archit saw right away. Lesson learned, (for about the 1000th time :p).
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@Brian Charlesworth – Meanwhile, Brian, can you do me a favor? Go solve this problem Brilli's faithfulness to... and legitimize it. I had the author put in the diagram, but I blew it anyway. I hadn't had my coffee yet. Once you put in the solution, then I can finally comment.
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@Michael Mendrin – I did get the answer on my second attempt. It was more difficult than I expected and the point P was not where I was initially expecting it to be. I'm still looking for a more elegant solution method before posting anything; did you want me to just post something short and incomplete so that you could post a comment/solution?
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@Brian Charlesworth – Yeah, something like that, and then I could expand on the fact that the correct solution is "over there, not here"
@Brian Charlesworth – Brian, how about if you (or Steven) do the honors and post a clever problem based on this?
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@Michael Mendrin – I would love to see what kind of devious thing @Michael Mendrin would come up with.
@Michael Mendrin – When I'm feeling remotely clever I'll see what I can come up with, but feel free to be "devious", as Steven suggests.
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@Brian Charlesworth – hey but I am no deviant!
all right let me go for a walk and deviant something up
same way :p jee style
I'm getting the same sum but with a different point. I get the plane equation as
< 4 , 5 , 6 > ∘ < x − 1 , y − 2 , z − 3 > = 0 ⟹ 4 x + 5 y + 6 z = 3 2 .
I then plugged in the same parametric equations as you have into this plane equation and found that t = 2 , giving
( a , b , c ) = ( − 3 , − 2 , 9 ) ⟹ a + b + c = 4 .
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I had (-3,-2,9) as well
Okay, I'll pull my solution until I find out what's going on.
Edit: Okay, is that better? It's kind of amazing that somehow I managed to get the same answer the other way. Was that some weird kind of duality?
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O.k., looks good now. It did seem like too much of a coincidence that your previous sum was 4 as well to not think that something interesting was going on. Your plane equation before seemed like a "reciprocal plane" with normal ( 4 1 , 5 1 , 6 1 ) . I can't see just yet why the same coordinate sum should necessarily pop out.
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@Brian Charlesworth – Yes, there's something here. Given an arbitrary normal vector which defines a plane perpendicular to it, passing through one arbitrary point, along with it the dual where the terms of normal vector are reciprocals of those in the first normal vector, etc., and then given an arbitrary point ( a , b , c ) , and (almost) arbitrary vector ( r , s , t ) , defining a line which intersects the two planes, yielding two points, then the sum of the coordinates of those two points are the same and are equal to the sum a + b + c ....IF the sum of the terms in the vector ( r , s , t ) equals zero, i.e., r + s + t = 0 . It's interesting how everything just drops right out. And I don't have an easy explanation for why this is, geometrically. Probably a great idea for either a note or a problem.
Notice that − 1 + 0 + 5 = 4 , and − 1 − 1 + 2 = 0 , in this problem. It was just a wild coincidence that Steven Chase picked the vector ( − 1 , − 1 , 2 ) which terms happens to add up to 0 . Otherwise, I would not have come up with the answer 4 for all the wrong reasons.
Edit: Oh, wait, I think I get it now. If the terms of that vector ( r , s , t ) sums up to 0 , then given any starting point ( a , b , c ) , the line so defined intersects any arbitrary plane at a point which coordinate terms have the same sum as a + b + c . There is no "duality" here, I could have used inverses of π , e , γ and it still would have worked out.
I don't think it's hard to prove this, but I'm going to sleep now.
Edit: It's morning now. See Archit Agrawa's brief comment above, which says it all.
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@Michael Mendrin – Yeah, Archit really did cut to the Chase, huh?
@Michael Mendrin – i think your final point is wrong,when you plug in t = 2 you get the point ( − 3 , − 2 , 9 )
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@Anirudh Sreekumar – Thanks for the catch, yes, I missed that one
The parametric equations of the line passing through ( − 1 , 0 , 5 ) in the direction of the vector ( − 1 , − 1 , 2 ) are
x = − 1 − t
y = 0 − t
z = 5 + 2 t
where t is the parameter. The plane equation is, if the normal vector is ( 4 , 5 , 6 ) and the plane includes the point ( 1 , 2 , 3 )
4 x + 5 y + 6 z − ( 4 ⋅ 1 + 5 ⋅ 2 + 6 ⋅ 3 ) = 0
Solve for t , which works out to t = 2 . Plug this back into the parametric equations and end up with
( x , y , z ) = ( − 3 , − 2 , 9 )
the sum of which is 4