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anonymous
 5 years ago
Why is the minimum distance from the plane x+2y+3z=6 to the origin,
\[6\div\sqrt{1^{2}+2^{2}+3^{2}}\]
and not just
\[\sqrt{1^{2}+2^{2}+3^{2}}\]
anonymous
 5 years ago
Why is the minimum distance from the plane x+2y+3z=6 to the origin, \[6\div\sqrt{1^{2}+2^{2}+3^{2}}\] and not just \[\sqrt{1^{2}+2^{2}+3^{2}}\]

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the equation of the plane involves the 6 as well

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the minimum distanceto the origin would be the point on the plane the is perped to the origin right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the normal vector appears to be <1,2,3> by looking at it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got the same normal vector btw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the modulus of the normal vector isnt necessarily the same vector that is pointing to the point from the origin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but direction doesnt matter when i square negative values

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its not the direction you need to determine really; its the nearest point to the origin shich isnt the same

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if you take the normal vector and scale it fromthe origin to meet the plane; that should do ti

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should i just construct a line that goes through the origin and plane with the normal vector (1,2,3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then find the distance when i find the point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this just seems to be a method thats a bit long, is there nothing simpler

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there prolly is a simpler method; but none that come to mind :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
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