anonymous
  • anonymous
Find an equation of the plane. The plane passes through the points (5, 3, 1) and (5, 1, -5) and is perpendicular to the plane 8x + 9y + 2z = 11.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I found the vector of two points, crossed points w/ a,b,c from eq of plane. then I plugged into eq. of plane formula. what did I do wrong?
anonymous
  • anonymous
Check your answer by taking dot product of two perpendicular vectors.
anonymous
  • anonymous
I see a mistake in my calculation.

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anonymous
  • anonymous
What is wrong in my method?
anonymous
  • anonymous
Show us your method.
anonymous
  • anonymous
(5,3,1) (5,1,-5) -> <0,-2,-6> <0,-2,-6>X<8,9,2>=<50,-48,16> <50,-48,16>.<8,9,2> = 0 <50,-48,16>=<25,-24,8> 25(x-5)^2-24(y-3)^2+8(z-1)=0 25x-125-24y-72+8z-8=0 25x-24y+8z=205
anonymous
  • anonymous
What does your answer say in book, apparently you are checking it?
myininaya
  • myininaya
chag (sorry to interrupt) did you ask about the calculus dream problem or was that someone else
anonymous
  • anonymous
Someone else.
myininaya
  • myininaya
k thanks
anonymous
  • anonymous
There is another method, I don't know if this is valid dot V=0. it gives you 25x-27y+2z=0
anonymous
  • anonymous
Not correct
anonymous
  • anonymous
You asked what is wrong with your method. The mistake is when you did the equation formula you squared x and y; you shouldn't have squared it 25x-24y+8z=? Plug in (5,3,1) and you get 25x-24y+8z=61

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