Calculating components of normal to a surface
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Let $S$ be the surface $−5x^2+4y^2+2z^2−4=−5$.
The vector $mathbf{n}$ is normal to $S$ at the point $(3,−3,2)$ and $mathbf{n}cdothat{mathbf{i}}=−30$. Find $mathbf{n}cdothat{mathbf{k}}$.
I found the normal to equal $(-30,-24,8)$ at the point $(3,-3,2)$. But I'm not sure what to do after that. What am I suppose to do with $mathbf{n}cdothat{mathbf{i}}=−30$? And how will that help me find $mathbf{n}cdothat{mathbf{k}}$ ?
vector-spaces vector-fields
edited Nov 25 at 17:21
B. Mehta
11.8k22144
asked Nov 25 at 17:15
Neels
227
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up vote
0
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Let $S$ be the surface $−5x^2+4y^2+2z^2−4=−5$.
The vector $mathbf{n}$ is normal to $S$ at the point $(3,−3,2)$ and $mathbf{n}cdothat{mathbf{i}}=−30$. Find $mathbf{n}cdothat{mathbf{k}}$.
I found the normal to equal $(-30,-24,8)$ at the point $(3,-3,2)$. But I'm not sure what to do after that. What am I suppose to do with $mathbf{n}cdothat{mathbf{i}}=−30$? And how will that help me find $mathbf{n}cdothat{mathbf{k}}$ ?
vector-spaces vector-fields
edited Nov 25 at 17:21
B. Mehta
11.8k22144
asked Nov 25 at 17:15
Neels
227
1
As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $S$ be the surface $−5x^2+4y^2+2z^2−4=−5$.
The vector $mathbf{n}$ is normal to $S$ at the point $(3,−3,2)$ and $mathbf{n}cdothat{mathbf{i}}=−30$. Find $mathbf{n}cdothat{mathbf{k}}$.
I found the normal to equal $(-30,-24,8)$ at the point $(3,-3,2)$. But I'm not sure what to do after that. What am I suppose to do with $mathbf{n}cdothat{mathbf{i}}=−30$? And how will that help me find $mathbf{n}cdothat{mathbf{k}}$ ?
vector-spaces vector-fields
edited Nov 25 at 17:21
B. Mehta
11.8k22144
asked Nov 25 at 17:15
Neels
227
Let $S$ be the surface $−5x^2+4y^2+2z^2−4=−5$.
The vector $mathbf{n}$ is normal to $S$ at the point $(3,−3,2)$ and $mathbf{n}cdothat{mathbf{i}}=−30$. Find $mathbf{n}cdothat{mathbf{k}}$.
I found the normal to equal $(-30,-24,8)$ at the point $(3,-3,2)$. But I'm not sure what to do after that. What am I suppose to do with $mathbf{n}cdothat{mathbf{i}}=−30$? And how will that help me find $mathbf{n}cdothat{mathbf{k}}$ ?
vector-spaces vector-fields
vector-spaces vector-fields
edited Nov 25 at 17:21
B. Mehta
11.8k22144
asked Nov 25 at 17:15
Neels
227
edited Nov 25 at 17:21
B. Mehta
11.8k22144
asked Nov 25 at 17:15
Neels
227
edited Nov 25 at 17:21
B. Mehta
11.8k22144
edited Nov 25 at 17:21
B. Mehta
11.8k22144
edited Nov 25 at 17:21
B. Mehta
11.8k22144
11.8k22144
asked Nov 25 at 17:15
Neels
227
asked Nov 25 at 17:15
Neels
227
asked Nov 25 at 17:15
Neels
227
227
1
As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26
add a comment |
1
As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26
1
1
As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26
As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26
add a comment |
1 Answer
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oldest
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up vote
2
down vote
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Assuming you have the correct normal vector (I haven't checked the first part of the calculation) then:
$vec{i}, vec{j}, vec{k}$ are the unit vectors in the x, y and z directions.
The question gives that $vec{n}cdotvec{i} = -30$, which is consistent with your computed value for $vec{n}$, i.e. it is the x component of your vector in Cartesian coordinates.
$vec{n}cdotvec{k}$ is the z component of your vector, which is 8.
answered Nov 25 at 17:30
ip6
54839
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Assuming you have the correct normal vector (I haven't checked the first part of the calculation) then:
$vec{i}, vec{j}, vec{k}$ are the unit vectors in the x, y and z directions.
The question gives that $vec{n}cdotvec{i} = -30$, which is consistent with your computed value for $vec{n}$, i.e. it is the x component of your vector in Cartesian coordinates.
$vec{n}cdotvec{k}$ is the z component of your vector, which is 8.
answered Nov 25 at 17:30
ip6
54839
add a comment |
up vote
2
down vote
accepted
Assuming you have the correct normal vector (I haven't checked the first part of the calculation) then:
$vec{i}, vec{j}, vec{k}$ are the unit vectors in the x, y and z directions.
The question gives that $vec{n}cdotvec{i} = -30$, which is consistent with your computed value for $vec{n}$, i.e. it is the x component of your vector in Cartesian coordinates.
$vec{n}cdotvec{k}$ is the z component of your vector, which is 8.
answered Nov 25 at 17:30
ip6
54839
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Assuming you have the correct normal vector (I haven't checked the first part of the calculation) then:
$vec{i}, vec{j}, vec{k}$ are the unit vectors in the x, y and z directions.
The question gives that $vec{n}cdotvec{i} = -30$, which is consistent with your computed value for $vec{n}$, i.e. it is the x component of your vector in Cartesian coordinates.
$vec{n}cdotvec{k}$ is the z component of your vector, which is 8.
answered Nov 25 at 17:30
ip6
54839
Assuming you have the correct normal vector (I haven't checked the first part of the calculation) then:
$vec{i}, vec{j}, vec{k}$ are the unit vectors in the x, y and z directions.
The question gives that $vec{n}cdotvec{i} = -30$, which is consistent with your computed value for $vec{n}$, i.e. it is the x component of your vector in Cartesian coordinates.
$vec{n}cdotvec{k}$ is the z component of your vector, which is 8.
answered Nov 25 at 17:30
ip6
54839
answered Nov 25 at 17:30
ip6
54839
answered Nov 25 at 17:30
ip6
54839
answered Nov 25 at 17:30
ip6
54839
54839
add a comment |
add a comment |
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As you can see, you were given the first component of that vector. $mathbf ncdotmathbf{hat k}$ is simply the third component. For the record, the required normal vector might also have been a multiple of the vector you found.
– I like Serena
Nov 25 at 17:26