I need to find two parallel planes that one of them go thru the points $\ (0,9,2),(-3,10,-1) $ and the second go thru $\ (2,2,2) , (-1,-1,5) $

So the direction vector of the first plane should be $\ v_1 = (0,9,2) - (-3,10,-1) = (3,-1,3) $ and the direction vector of the second one is $\ v_2 =(3,3,-3) $

now since the planes are parallel I can use the cross product of those two directional vectors to get the perpendicular vector to those planes (am I correct) ? If so, the cross product of those two:

$\ v_1 \times v_2 = -6i - 18j + 12z = (-1,-3,2) $

and having the perpendicular vector I know that according to this answer the dot product of any vector in those planes would gives zero hence $\ 0 = ((x,y,z)-(0,9,2))\cdot(-1,-3,2) $ which gives the end result of $\ 0 = -x-3y+2z+23 $ for the first plane

but looking ploting this plane in a graph it doesn't seems like the point are on this plane?

I guess i've misundestood something here?


You have a typo in the second component of cross product - it should have the opposite sign. Then for the plane you get equation $-x + 3y + 2z - 31 = 0$, and both points satisfy this equation.


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