Equation of plane through point P containing line L

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SUMMARY

The discussion focuses on finding the equation of a plane W2 that passes through the point P(2,3,-1) and contains the line L defined by the parametric equation L = (3,-1,0) + t(5/3, -1/3, 1). The correct points on the line for t = 3 and t = 6 are identified as Q(8, -2, 3) and R(13, -3, 6), respectively. The normal vector to the plane is calculated using the cross product of vectors PQ and PR, resulting in the equation 14x - 7y + 21z + 14 = 0. The discussion emphasizes the importance of accurate calculations and understanding vector operations in determining the plane's equation.

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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
 
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phyzmatix said:
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:
dynamicsolo said:
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:
phyzmatix said:
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...).
 
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!
 
cemar. said:
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 PQ \times PR. 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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