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the equation of the plane involves the 6 as well
x+2y +3z -6 = 0
the minimum distanceto the origin would be the point on the plane the is perped to the origin right?
the normal vector appears to be <1,2,3> by looking at it
yeah as far as i understand. But i don't get why you can't just take the modulus of the normal vector? where does the 6 come in?
i got the same normal vector btw
the modulus of the normal vector isnt necessarily the same vector that is pointing to the point from the origin
but direction doesnt matter when i square negative values
its not the direction you need to determine really; its the nearest point to the origin shich isnt the same
if you take the normal vector and scale it fromthe origin to meet the plane; that should do ti
should i just construct a line that goes through the origin and plane with the normal vector (1,2,3)
then find the distance when i find the point
i would yes
this just seems to be a method thats a bit long, is there nothing simpler
there prolly is a simpler method; but none that come to mind :)
the distance from a point P1(x1,y1,z1) to th plane: ax+by+cz+d=0 is: D= /ax1+by1+cz1+d/ : sq.rt(a^2 + b^2+c^2) I can show you solution in full if you need it