anonymous
  • anonymous
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 \[\sqrt{1^{2}+2^{2}+3^{2}}\]
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amistre64
  • amistre64
the equation of the plane involves the 6 as well
amistre64
  • amistre64
x+2y +3z -6 = 0
amistre64
  • amistre64
the minimum distanceto the origin would be the point on the plane the is perped to the origin right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
the normal vector appears to be <1,2,3> by looking at it
anonymous
  • 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?
anonymous
  • anonymous
i got the same normal vector btw
amistre64
  • amistre64
the modulus of the normal vector isnt necessarily the same vector that is pointing to the point from the origin
anonymous
  • anonymous
but direction doesnt matter when i square negative values
amistre64
  • amistre64
its not the direction you need to determine really; its the nearest point to the origin shich isnt the same
amistre64
  • amistre64
if you take the normal vector and scale it fromthe origin to meet the plane; that should do ti
anonymous
  • anonymous
should i just construct a line that goes through the origin and plane with the normal vector (1,2,3)
anonymous
  • anonymous
then find the distance when i find the point
amistre64
  • amistre64
i would yes
anonymous
  • anonymous
this just seems to be a method thats a bit long, is there nothing simpler
amistre64
  • amistre64
there prolly is a simpler method; but none that come to mind :)
anonymous
  • anonymous
thanks nyways
amistre64
  • amistre64
yw
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.