Consider the solution set S of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$ . Calculate the...
$begingroup$
Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.
linear-algebra
edited Jan 14 at 0:00
Bernard
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
$endgroup$
add a comment |
$begingroup$
Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.
linear-algebra
edited Jan 14 at 0:00
Bernard
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
$endgroup$
$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09
add a comment |
$begingroup$
Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.
linear-algebra
edited Jan 14 at 0:00
Bernard
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
$endgroup$
Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.
linear-algebra
linear-algebra
edited Jan 14 at 0:00
Bernard
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
edited Jan 14 at 0:00
Bernard
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
edited Jan 14 at 0:00
Bernard
124k742117
edited Jan 14 at 0:00
Bernard
124k742117
edited Jan 14 at 0:00
Bernard
124k742117
124k742117
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
asked Jan 13 at 23:42
hessel roodenburghessel roodenburg
31
31
$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09
add a comment |
$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09
$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane
In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$
edited Jan 14 at 0:19
Bernard
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
$endgroup$
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
add a comment |
$begingroup$
Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:
Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$
The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$
Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$
answered Jan 14 at 0:45
BernardBernard
124k742117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane
In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$
edited Jan 14 at 0:19
Bernard
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
$endgroup$
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
add a comment |
$begingroup$
It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane
In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$
edited Jan 14 at 0:19
Bernard
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
$endgroup$
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
add a comment |
$begingroup$
It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane
In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$
edited Jan 14 at 0:19
Bernard
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
$endgroup$
It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane
In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$
edited Jan 14 at 0:19
Bernard
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
edited Jan 14 at 0:19
Bernard
124k742117
edited Jan 14 at 0:19
Bernard
124k742117
edited Jan 14 at 0:19
Bernard
124k742117
124k742117
answered Jan 14 at 0:10
user289143user289143
1,069313
answered Jan 14 at 0:10
user289143user289143
1,069313
answered Jan 14 at 0:10
user289143user289143
1,069313
1,069313
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
add a comment |
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
$begingroup$
But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
$endgroup$
– hessel roodenburg
Jan 14 at 0:20
1
1
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
$begingroup$
The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
$endgroup$
– user289143
Jan 14 at 0:23
add a comment |
$begingroup$
Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:
Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$
The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$
Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$
answered Jan 14 at 0:45
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:
Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$
The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$
Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$
answered Jan 14 at 0:45
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:
Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$
The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$
Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$
answered Jan 14 at 0:45
BernardBernard
124k742117
$endgroup$
Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:
Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$
The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$
Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$
answered Jan 14 at 0:45
BernardBernard
124k742117
answered Jan 14 at 0:45
BernardBernard
124k742117
answered Jan 14 at 0:45
BernardBernard
124k742117
answered Jan 14 at 0:45
BernardBernard
124k742117
124k742117
add a comment |
add a comment |
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$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03
$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09