SOLUTION: Let u=i-2j+4k Find the equation of the plane that has a normal u and contains a point (2,3,4). Thanks, Judy

Question 227981: Let u=i-2j+4k
Find the equation of the plane that has a normal u and contains a point (2,3,4).
Thanks, Judy
Found 2 solutions by Alan3354, user_dude2008:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Let u=i-2j+4k
Find the equation of the plane that has a normal u and contains a point (2,3,4).
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Parallel planes normal to u:
x - 2y + 4z = k
-----------------
Sub point
2 - 2*3 + 4*4 = k = 2 -6 +8
k = 4
------------
x - 2y + 4z = 4

Answer by user_dude2008(1862) (Show Source): You can put this solution on YOUR website!
equation of plane: ax+by+cz=d ----> is normal vector

Plug in normal vector u=i-2j+4k: ax+by+cz=d ---> x-2y+4z=d

Plug in point (x,y,z)=(2,3,4) ----> 2-2(3)+4(4)=12=d

equation of plane is x-2y+4z=12
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