anonymous
  • anonymous
Find a vector perpendicular to i +2j -k and 3i-j+k. ---> i - 4j -7k (I know how to get this) Hence, find the cartesian equation of the plane parallel to both i +2j-k and 3i-j+k which passes through the point (2,0,-3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@myko
anonymous
  • anonymous
(x,y,z)= (2,0,-2)+a(1,2-1)+b(3,-1,1)
anonymous
  • anonymous
x-2=a+b y = 2a-b z+2 = -a +b eliminate a and b from here and you get the equation

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anonymous
  • anonymous
ok?
anonymous
  • anonymous
How to eliminate a and b?
anonymous
  • anonymous
adding for example 1º and 3º equation you get: x-2 +z+2 = 2b so b = (x-2 +z +2 )/2 adding 1º and 2º: x-2 +y = 3a so a = (x-2 +y )/3 Now put this values in some equation and put it in the form Ax+By+Cz = D
anonymous
  • anonymous
ok?
anonymous
  • anonymous
i think if i didn't make any mistakes: x-2y-3z =8 check it just in case...
anonymous
  • anonymous
ups, i found mistake: x-2=a+b y = 2a-b z+2 = -a +b should be: x-2=a+3b !!!!!!!! y = 2a-b z+2 = -a +b so in the rest fix this value
anonymous
  • anonymous
result is x-4y-7z =-16 this should be right
anonymous
  • anonymous
got it?
anonymous
  • anonymous
No.. Not really.... You should get at the end. x-4y-7z=23
anonymous
  • anonymous
impossible, doesn't contain the point (2,0,-2)
anonymous
  • anonymous
That's what it says
anonymous
  • anonymous
try it your self. Put (2,0,-2) in the equation x-4y-7z=23 and it doesn't match it
anonymous
  • anonymous
Oh, I made a type. It's meant to be (2, 0, -3)
anonymous
  • anonymous
damm...:)
anonymous
  • anonymous
ya, than it gives x-4y-7z=23
anonymous
  • anonymous
@order
anonymous
  • anonymous
How?
anonymous
  • anonymous
x-2=a+3b y = 2a-b z+3 = -a +b from here: a =y+z+3, b = (x-2+z+3)/4 puting this values in the last equation: z+3=-y-z-3+ (x-2+z+3)/4 and regrouping gives x-4y-7z=23

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