
#1
Jun1808, 12:55 PM

P: 364

1. The problem statement, all variables and given/known data
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 



#2
Jun1808, 06:01 PM

HW Helper
P: 1,664

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, 11, 0+3) = (8, 2, 3) and R: (3,1,0) + 6(5/3, 1/3, 1) = (3+10, 12, 0+6) = (13, 3, 6) , shouldn't they? Otherwise, in principle, your method looks all right... 



#3
Jun1908, 01:56 AM

P: 364

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 



#4
Jun1908, 10:39 AM

HW Helper
P: 1,664

Equation of plane through point P containing line L 



#5
Dec209, 12:56 AM

P: 41

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
Dec209, 04:41 AM

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