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

$$\sqrt{1^{2}+2^{2}+3^{2}}$$

Question

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

$$\sqrt{1^{2}+2^{2}+3^{2}}$$

amistre64 · Answer

the equation of the plane involves the 6 as well

amistre64 · Answer

x+2y +3z -6 = 0

amistre64 · Answer

the minimum distanceto the origin would be the point on the plane the is perped to the origin right?

amistre64 · Answer

the normal vector appears to be <1,2,3> by looking at it

anonymous · Answer

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?

anonymous · Answer

i got the same normal vector btw

amistre64 · Answer

the modulus of the normal vector isnt necessarily the same vector that is pointing to the point from the origin

anonymous · Answer

but direction doesnt matter when i square negative values

amistre64 · Answer

its not the direction you need to determine really; its the nearest point to the origin shich isnt the same

amistre64 · Answer

if you take the normal vector and scale it fromthe origin to meet the plane; that should do ti

anonymous · Answer

should i just construct  a line that goes through the origin and plane with the normal vector (1,2,3)

anonymous · Answer

then find the distance when i find the point

amistre64 · Answer

i would yes

anonymous · Answer

this just seems to be a method thats a bit long, is there nothing simpler

amistre64 · Answer

there prolly is a simpler method; but none that come to mind :)

anonymous · Answer

thanks nyways

amistre64 · Answer

yw

anonymous · Answer

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