Jtdylktuy

I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.

The following is my work:

$$d = sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$

since $x+y+z = 6$, $z = 6-x-y$, so

begin{align*}
d &= sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
end{align*}

Find partial derivative $f_x$ and $f_y$ and critical points

begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \
&= 24-2y quad (text{set }= 0) \
&= text{critical point }y = 4 \
f_y &= 2y + 2(-x-y+12) \
&= 24 - 2x quad (text{set }= 0) \
&= text{critical point }x = 12 \
end{align*}

Plug in $x = 12$ and $y = 4$ to original equation

$$d = sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = sqrt{48}$$

Your approach is a good one, and in fact can lead to the general formula mentioned in the other answers.

You just made a few mistakes in your calculation.

For example, you say that $frac{partial}{partial x} left[ (x-8)^2+y^2+(-x-y+12)^2 right]$ is $2(x-8)+2(-x-y+12)$, when really it is $2(x-8)-2(-x-y-12)$. You forgot the chain rule.

You make a similar chain rule mistake when calculating the partial derivative with respect to $y$.

I think if you fix these, you should be able to finish your calculation the way you started it.

Note to other posters: I think it is generally of greater pedagogical value to help a student realize how to fix their own reasoning, rather than supplying them with a totally different method. Our primary aim should be to empower students to realize that they have the power to think through these things logically and creatively.

The distance formula is $$D=frac{|ax_0+by_0 +cz_0-d|}{sqrt{a^2+b^2+c^2}}=frac{4}{sqrt3}.$$

The closest distance will be along a line perpendicular to the plane or parallel to $(1,1,1)$. Since $8+0+-6=2$ we must add $frac{4}{3}$ to each component so that $x+y+z$ now equals $6$. Therefore, the distance is $|(frac{4}{3}, frac{4}{3}, frac{4}{3})|=sqrt{3frac{16}{9}}=sqrt{frac{16}{3}}=frac{4}{sqrt{3}}$

It's $$frac{|8+0-6-6|}{sqrt{1^2+1^2+1^2}}=frac{4}{sqrt3}.$$
I used the following formula.

The distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ it's
$$frac{|ax_1+by_1+cz_1+d|}{sqrt{a^2+b^2+c^2}}.$$

Also, we can end your way.

Indeed, by C-S
$$sqrt{(x-8)^2+y^2+(z+6)^2}=frac{1}{sqrt3}sqrt{(1^2+1^2+1^2)((x-8)^2+y^2+(z+6)^2)}geq$$
$$geqfrac{1}{sqrt3}sqrt{(x-8+y+z+6)^2}=frac{4}{sqrt3}$$ and we got the distance again.

Option:

Normal of the plane: $(1,1,1)$,

Normalized(i.e of unit length): $(1/√3)(1,1,1)$

Line through $(8,0,-6)$:

$vec r= (8,0,-6)+t (1/√3)(1,1,1)$.

Determine $t$ when line intersects plane:

$(8+t/√3)+t/√3+(-6+t/√3)=6;$

$3(t/√3)=4$;

$t= 4/√3.$

Distance = $4/√3$ (Why?).

You can draw a line from your point to the plane and then find the length of its projection onto the unit normal vector to the plane.