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How to find equation of the plane with three coordinates?
Find the cartesian equation of the plane which passes through the points:
A(3,0,0) B(2,0,5) and C(4,-3,1)
Now googling this to find some answers and I came across this cross product thing, this is something I do not do for my MEI maths C4 exam in two days, I have to use another method.
Normally if I am given the normal vector to the plane which happens in most cases, I can simply use: n.r = n.a
However there are sometimes where I do not get given the normal and only three coordinates, I've never seen it come up in the exam, only in my textbook. I would really like some help trying to figure this out. In my book it has something to do with simultaneous equations however everytime I try it I get the wrong answer.
Thank you for your time.
2 Answers
- Let'squestionLv 78 years agoFavorite Answer
As the plane passes through A, its equation must be of the form
l(x-3) + m(y-0) + n(z-0) = 0 ---------------------- (1) As it passes through B we have
l*(2-3)+ m(0-0) +n*(5-0) = 0 or
-l +5n = 0 ------------------- (2), silarly as it passes through C we have
l*(4-3)+ m(-3-0) +n*(1-0) = 0 or
l -3m +n = 0 -------------------(3); Adding (2) and (3) we get
-3m +6n = 0 or m = 2n ---------------- (4) Substituting in (3) we get
l = 5n--------------------- (5) substituting (4) and (5) in (10 we get
5n(x-3) +2ny +nz = 0 or 5x -15 +2y +z = 0 or
The required equation is
5x + 2y + z = 15; You can verify by substituting coordinates of A B and C.
- ?Lv 45 years ago
for this question you should use vector calculus. you may discover 2 vectors, probable ab and ac, or inspite of, then take the go product, then use that to outline your airplane. v1= i + j + 2k v2= -2i + 0j + 2k v1 X v2 = 2i - 6j + 2k fill on your planar equation using this go product (perpendicular to the wanted airplane) and between the coordinates 2(x-0) -6(y-a million) + 2(z+a million)=0 2x -6y + 2z = -6 - 2 = -8 x - 3y + z = -4 is your very last planar equation :)