If I remember correctly and didn't make arithmetic errors
(P.S. English is not my native)
1)Make the equation of the plane by the point and two non-collinear vectors:
|x - 1 1 -1 |
|y - 2 -3 1 | = 0
|z-(-2) 1 2 |
(x-1)*|-3 1| - (y-2)*|1 -1| +(z+2)*|1 -1|= 0
|1 2 | |1 2| |-3 1|
(x-1)*(-6-2) - (y-2)*(2+1)+(z+2)*(1-3)=0
-8x+8 -3y + 6 -2z-4 =0
-8x -3y -2z +10=0
8x +3y +2z -10 =0
2)Make the equation to the normal form
To find the normal equation of the plane defined by the equation Ax + By + Cz + D = 0, it suffices to divide both sides of this equation by:
where the upper sign is taken when D> 0, and lower when D <0; if D = 0, then we can take any sign.
So divide both sides by:
____________ __
\/8*8+3*3+2*2 = \/77
8___ x + 3 ___ y + 2___ z - 10___ = 0
Answer:\/77 \/77 \/77 \/77
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