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

1. amistre64

the equation of the plane involves the 6 as well

2. amistre64

x+2y +3z -6 = 0

3. amistre64

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

4. amistre64

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

5. anonymous

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?

6. anonymous

i got the same normal vector btw

7. amistre64

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

8. anonymous

but direction doesnt matter when i square negative values

9. amistre64

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

10. amistre64

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

11. anonymous

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

12. anonymous

then find the distance when i find the point

13. amistre64

i would yes

14. anonymous

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

15. amistre64

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

16. anonymous

thanks nyways

17. amistre64

yw

18. anonymous

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