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2

$begingroup$

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

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

multivariable-calculus optimization cauchy-schwarz-inequality

share|cite|improve this question

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did the $6$ become the $12$?
  $endgroup$
  – Theo Bendit
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because z = - x - y + 6. So since I plug in (z + 6)^2 in the distance formula, it becomes (-x - y + 12)^2
  $endgroup$
  – Niko
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

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

multivariable-calculus optimization cauchy-schwarz-inequality

share|cite|improve this question

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did the $6$ become the $12$?
  $endgroup$
  – Theo Bendit
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because z = - x - y + 6. So since I plug in (z + 6)^2 in the distance formula, it becomes (-x - y + 12)^2
  $endgroup$
  – Niko
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

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

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

multivariable-calculus optimization cauchy-schwarz-inequality

share|cite|improve this question

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

$endgroup$

share|cite|improve this question

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

share|cite|improve this question

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

edited 6 hours ago

Michael Rozenberg

105k1892197

edited 6 hours ago

Michael Rozenberg

105k1892197

asked 7 hours ago

NikoNiko

132

asked 7 hours ago

NikoNiko

132

5 Answers
5

active

oldest

votes

2

$begingroup$

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

share|cite|improve this answer

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

add a comment | 

2

$begingroup$

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 $|(frac{4}{3}, frac{4}{3}, frac{4}{3})|=sqrt{3frac{16}{9}}=sqrt{frac{16}{3}}=frac{4}{sqrt{3}}$

share|cite|improve this answer

answered 6 hours ago

John DoumaJohn Douma

5,53211319

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much! this I smuch quicker than what I am being told to do
  $endgroup$
  – Niko
  7 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

BootsBoots

15410

$endgroup$

add a comment | 

0

$begingroup$

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?).

share|cite|improve this answer

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

$endgroup$

add a comment | 

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2

$begingroup$

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

share|cite|improve this answer

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this answer

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

answered 7 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

2

$begingroup$

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 $|(frac{4}{3}, frac{4}{3}, frac{4}{3})|=sqrt{3frac{16}{9}}=sqrt{frac{16}{3}}=frac{4}{sqrt{3}}$

share|cite|improve this answer

answered 6 hours ago

John DoumaJohn Douma

5,53211319

$endgroup$

add a comment | 

2

$begingroup$

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 $|(frac{4}{3}, frac{4}{3}, frac{4}{3})|=sqrt{3frac{16}{9}}=sqrt{frac{16}{3}}=frac{4}{sqrt{3}}$

share|cite|improve this answer

answered 6 hours ago

John DoumaJohn Douma

5,53211319

$endgroup$

add a comment | 

$begingroup$

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 $|(frac{4}{3}, frac{4}{3}, frac{4}{3})|=sqrt{3frac{16}{9}}=sqrt{frac{16}{3}}=frac{4}{sqrt{3}}$

share|cite|improve this answer

answered 6 hours ago

John DoumaJohn Douma

5,53211319

$endgroup$

share|cite|improve this answer

answered 6 hours ago

John DoumaJohn Douma

5,53211319

answered 6 hours ago

John DoumaJohn Douma

5,53211319

answered 6 hours ago

John DoumaJohn Douma

5,53211319

1

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much! this I smuch quicker than what I am being told to do
  $endgroup$
  – Niko
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much! this I smuch quicker than what I am being told to do
  $endgroup$
  – Niko
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

answered 7 hours ago

Michael RozenbergMichael Rozenberg

105k1892197

0

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

BootsBoots

15410

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

BootsBoots

15410

$endgroup$

add a comment | 

$begingroup$

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.

share|cite|improve this answer

answered 6 hours ago

BootsBoots

15410

$endgroup$

share|cite|improve this answer

answered 6 hours ago

BootsBoots

15410

answered 6 hours ago

BootsBoots

15410

answered 6 hours ago

BootsBoots

15410

0

$begingroup$

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?).

share|cite|improve this answer

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

$endgroup$

add a comment | 

0

$begingroup$

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?).

share|cite|improve this answer

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

$endgroup$

add a comment | 

$begingroup$

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?).

share|cite|improve this answer

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822

answered 4 hours ago

Peter SzilasPeter Szilas

11.4k2822