Mixing
Distance from a point to a plane $pi$
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Let $x=(1,2,3)$, $u=(-2,3,1)$. let $pi$ the plane through the origin which is orthogonal to the vector $u$.
Compute the distance from $x$ to $pi$.
The formula is: $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ .
Initially, I need to find a equation from point of $x$ to point of $u$, right?
How can I compute, can you help?
linear-algebra analytic-geometry
asked Jan 7 at 8:47
KathySongKathySong
1,270923
$endgroup$
add a comment |
$begingroup$
Let $x=(1,2,3)$, $u=(-2,3,1)$. let $pi$ the plane through the origin which is orthogonal to the vector $u$.
Compute the distance from $x$ to $pi$.
The formula is: $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ .
Initially, I need to find a equation from point of $x$ to point of $u$, right?
How can I compute, can you help?
linear-algebra analytic-geometry
asked Jan 7 at 8:47
KathySongKathySong
1,270923
$endgroup$
add a comment |
$begingroup$
Let $x=(1,2,3)$, $u=(-2,3,1)$. let $pi$ the plane through the origin which is orthogonal to the vector $u$.
Compute the distance from $x$ to $pi$.
The formula is: $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ .
Initially, I need to find a equation from point of $x$ to point of $u$, right?
How can I compute, can you help?
linear-algebra analytic-geometry
asked Jan 7 at 8:47
KathySongKathySong
1,270923
$endgroup$
Let $x=(1,2,3)$, $u=(-2,3,1)$. let $pi$ the plane through the origin which is orthogonal to the vector $u$.
Compute the distance from $x$ to $pi$.
The formula is: $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ .
Initially, I need to find a equation from point of $x$ to point of $u$, right?
How can I compute, can you help?
linear-algebra analytic-geometry
linear-algebra analytic-geometry
asked Jan 7 at 8:47
KathySongKathySong
1,270923
asked Jan 7 at 8:47
KathySongKathySong
1,270923
edited Jan 7 at 8:52
KathySong
asked Jan 7 at 8:47
KathySongKathySong
1,270923
asked Jan 7 at 8:47
KathySongKathySong
1,270923
asked Jan 7 at 8:47
KathySongKathySong
1,270923
1,270923
add a comment |
add a comment |
2 Answers
2
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oldest
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Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
add a comment |
$begingroup$
In the formula $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $pi,$ and $D=0$ because $pi$ passes through origin. Thus $$d=dfrac {left| 1times(-2)+2times3+3times1right| }{sqrt {(-2)^{2}+3^{2}+1^{2}}}=frac{7}{sqrt{14}}=frac{sqrt{14}}{2}$$
answered Jan 7 at 10:07
user376343user376343
3,3933826
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
add a comment |
$begingroup$
Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
add a comment |
$begingroup$
Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
$endgroup$
Given a normal vector $u=(a, b, c)$ for the plane and a point $O(x_0, y_0, z_0)$ on the plane, then the plane equation is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Your $u=(-2, 3, 1)$ and $O=(0, 0, 0)$ so the plane equation is $-2x+3y+z=0$. Now get the coefficients $A, B, C, D$ from here and plug $x$ in your distance formula to find the answer.
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
edited Jan 7 at 8:56
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
answered Jan 7 at 8:54
EuxhenHEuxhenH
484210
484210
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
add a comment |
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 8:56
add a comment |
$begingroup$
In the formula $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $pi,$ and $D=0$ because $pi$ passes through origin. Thus $$d=dfrac {left| 1times(-2)+2times3+3times1right| }{sqrt {(-2)^{2}+3^{2}+1^{2}}}=frac{7}{sqrt{14}}=frac{sqrt{14}}{2}$$
answered Jan 7 at 10:07
user376343user376343
3,3933826
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
add a comment |
$begingroup$
In the formula $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $pi,$ and $D=0$ because $pi$ passes through origin. Thus $$d=dfrac {left| 1times(-2)+2times3+3times1right| }{sqrt {(-2)^{2}+3^{2}+1^{2}}}=frac{7}{sqrt{14}}=frac{sqrt{14}}{2}$$
answered Jan 7 at 10:07
user376343user376343
3,3933826
$endgroup$
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
add a comment |
$begingroup$
In the formula $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $pi,$ and $D=0$ because $pi$ passes through origin. Thus $$d=dfrac {left| 1times(-2)+2times3+3times1right| }{sqrt {(-2)^{2}+3^{2}+1^{2}}}=frac{7}{sqrt{14}}=frac{sqrt{14}}{2}$$
answered Jan 7 at 10:07
user376343user376343
3,3933826
$endgroup$
In the formula $d=dfrac {left| Ax_{1}+Bx_{2}+Cx_{3}+Dright| }{sqrt {A^{2}+B^{2}+C^{2}}}$ are $A,B,C$ the coordinates of $u$ since this vector is orthogonal to $pi,$ and $D=0$ because $pi$ passes through origin. Thus $$d=dfrac {left| 1times(-2)+2times3+3times1right| }{sqrt {(-2)^{2}+3^{2}+1^{2}}}=frac{7}{sqrt{14}}=frac{sqrt{14}}{2}$$
answered Jan 7 at 10:07
user376343user376343
3,3933826
answered Jan 7 at 10:07
user376343user376343
3,3933826
answered Jan 7 at 10:07
user376343user376343
3,3933826
answered Jan 7 at 10:07
user376343user376343
3,3933826
3,3933826
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
add a comment |
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
$begingroup$
Thanks for answer.
$endgroup$
– KathySong
Jan 7 at 16:57
add a comment |
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