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Question 1143646: The equation of the plane passing thru points (5,4,1) ,(4,-2,-3) and (0,6,5) is expressed as x/A+y/B+z/C=1
Find the value of B
Find the value of C
Find the equation of the plane.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The equation of the plane passing thru points Q(5,4,1), R(4,-2,-3) and S(0,6,5) is expressed as x/A+y/B+z/C=1
Find the value of B
Find the value of C
Find the equation of the plane.
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Step 1, define 2 vectors, QR and RS.
Q - R = (1,6,4)
QR = i + 6j + 4k
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R - S = (4,-8,-8)
RS = 4i - 8j - 8k
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Step 2 find the cross product of QR and RS
|i j k|
|1 6 4| = i*(-48+32) - j*(-8-16) + k*(-8-24) = -16i + 24j - 32k
|4 -8 -8|
This the vector normal to the plane.
Since we're not concerned with its magnitude, reduce it to
2i - 3j + 4k
Call it N.
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Derive the equation of the plane.
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The equation of the plane is Ni(x-a1) + Nj(y-a2) + Nk(z-a3) = 0, where (a1,a2,a3) is any point in the plane and (Ni,Nj,Nk) is the normal vector, N.
In the example, using point Q, which is (5,4,1), the equation of the plane is 2(x-5) - 3(y-4) + 4(z-1) = 0, which simplifies to
2x-10 - 3y+12 + 4z-4 = 0, or
2x - 3y + 4z = 2
Verify your answer. Substitute the original points to see if they satisfy the equation of the plane.
(5,4,1)
2*5 - 3*4 + 4 = 2 OK
(4,-2-3)
2*4 + -3*-2 + 4*-3 = 2 OK
(0,6,5)
2*0 - 3*6 + 4*5 = 2 OK
The 3 points are in the plane.
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To express it in the form x/A+y/B+z/C=1 divide by 2.
x/2 + y/(-2/3) + z/(1/2) = 1
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Find the value of B
B = -2/3
Find the value of C
c = 1/2
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PS I made an Excel sheet that does these, if you're interested.
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