Equation of plane through point P containing line L

  • Thread starter phyzmatix
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  • #1
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Homework Statement



Find an equation for the plane W2 through the point P(2,3,-1) containing L

[L = (3,-1,0) + t(5/3, -1/3, 1)]


2. The attempt at a solution

Let Q and R be two points on L where t = 3 and t = 6 respectively

(t = 3) Q(11, -3, 3)
(t = 6) R(16, -4, 6)

Then PQ = (9, -6, 4) and PR = (14, -7, 7)

The normal to the plane W2 is then

PQ x PR = (14, -7, 21)

Finally

W2 : a(x - x0) + b(y - y0) + c(z - z0) = 0
14(x - 2) - 7(y - 3) + 21(z - +1) = 0
14x -7y +21z + 14 = 0

Could someone please be so kind as to check my calculations (especially for the two points Q and R, I'm not sure if I derived them correctly from the line's equation)? I'm not sure if my answer is correct since I can't find any similar examples in my textbook so I'd just like to know if my reasoning is sound.

Thanks in advance.

phyz
 

Answers and Replies

  • #2
dynamicsolo
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L = (3,-1,0) + t(5/3, -1/3, 1) ...


Let Q and R be two points on L where t = 3 and t = 6 respectively

(t = 3) Q(11, -3, 3)
(t = 6) R(16, -4, 6)


I understand why you chose t = 3 and t = 6 (though why not use t = 0?). But the values should be

Q: (3,-1,0) + 3(5/3, -1/3, 1) = (3+5, -1-1, 0+3) = (8, -2, 3) and
R: (3,-1,0) + 6(5/3, -1/3, 1) = (3+10, -1-2, 0+6) = (13, -3, 6) ,

shouldn't they?

Otherwise, in principle, your method looks all right...
 
Last edited:
  • #3
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I understand why you chose t = 3 and t = 6 (though why not use t = 0?). But the values should be

Q: (3,-1,0) + 3(5/3, -1/3, 1) = (3+5, -1-1, 0+3) = (8, -2, 3) and
R: (3,-1,0) + 6(5/3, -1/3, 1) = (3+10, -1-2, 0+6) = (13, -3, 6) ,

shouldn't they?

Otherwise, in principle, your method looks all right...
Thanks dynamicsolo! I just had a look here and I honestly can't tell you how I got to those values :confused:

Just goes to show that you shouldn't attempt mathematics at the end of the day when the old grey matter is starting to rebel against work :biggrin:
 
Last edited:
  • #4
dynamicsolo
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Just goes to show that you shouldn't attempt mathematics at the end of the day when the old grey matter is starting to rebel against work :biggrin:
I tell the students I work with that mathematical reasoning engages many of the so-called "higher brain functions" (there is also a lot of interaction going on between various sections of the brain). Those functions are the first thing to go when you're tired or hungry, so you become prone to increasing numbers of mistakes (sometimes bizarre ones -- I've written stuff at 2 AM that made next to no sense after a night's sleep...).
 
  • #5
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I am facing a similar problem right now, and i understand most of it, however, i am just having a hard time seeing how PQ * PR = normal of the plane
Thank you!
 
  • #6
HallsofIvy
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I am facing a similar problem right now, and i understand most of it, however, i am just having a hard time seeing how PQ * PR = normal of the plane
Thank you!
What do you mean by "PQ*PR"? It is the cross product of two vectors that is perpendicular to the plane containing the vectors and the standard notation for cross product is [itex]PQ \times PR[/itex]. And, of course, it is normal to the plane because the two vectors are in the plane and, as I said, "the cross product of two vectors is perpendicular to the plane containing them".
 

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