Dot product with a constant
$begingroup$
What is $vec{a}cdot (vec{b}+c)$, where $c$ is a constant?
Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
abstract-algebra vectors
edited 4 hours ago
greedoid
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is $vec{a}cdot (vec{b}+c)$, where $c$ is a constant?
Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
abstract-algebra vectors
edited 4 hours ago
greedoid
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago
add a comment |
$begingroup$
What is $vec{a}cdot (vec{b}+c)$, where $c$ is a constant?
Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
abstract-algebra vectors
edited 4 hours ago
greedoid
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
What is $vec{a}cdot (vec{b}+c)$, where $c$ is a constant?
Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
abstract-algebra vectors
abstract-algebra vectors
edited 4 hours ago
greedoid
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
greedoid
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
greedoid
43.5k1154107
edited 4 hours ago
greedoid
43.5k1154107
edited 4 hours ago
greedoid
43.5k1154107
43.5k1154107
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
asked 4 hours ago
ItIsLastThursdayItIsLastThursday
212
212
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ItIsLastThursday is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago
add a comment |
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.
$c$ must be a vector, then the sum is defined and the given product is a dot product.
edited 59 mins ago
Eevee Trainer
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
$endgroup$
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
add a comment |
$begingroup$
For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $vec b + c$ is not defined, and hence the expression $vec a cdot (vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.
Although it probably makes sense to you that
$$vec a cdot (vec b + c) = vec a cdot vec b + c cdot vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
$endgroup$
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
ItIsLastThursday is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3121781%2fdot-product-with-a-constant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.
$c$ must be a vector, then the sum is defined and the given product is a dot product.
edited 59 mins ago
Eevee Trainer
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
$endgroup$
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
add a comment |
$begingroup$
This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.
$c$ must be a vector, then the sum is defined and the given product is a dot product.
edited 59 mins ago
Eevee Trainer
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
$endgroup$
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
add a comment |
$begingroup$
This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.
$c$ must be a vector, then the sum is defined and the given product is a dot product.
edited 59 mins ago
Eevee Trainer
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
$endgroup$
This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.
$c$ must be a vector, then the sum is defined and the given product is a dot product.
edited 59 mins ago
Eevee Trainer
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
edited 59 mins ago
Eevee Trainer
6,53811237
edited 59 mins ago
Eevee Trainer
6,53811237
edited 59 mins ago
Eevee Trainer
6,53811237
6,53811237
answered 4 hours ago
greedoidgreedoid
43.5k1154107
answered 4 hours ago
greedoidgreedoid
43.5k1154107
answered 4 hours ago
greedoidgreedoid
43.5k1154107
43.5k1154107
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
add a comment |
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
$begingroup$
So is there any answer I can get?
$endgroup$
– ItIsLastThursday
4 hours ago
6
6
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
No, if $c$ is a scalar then there is no answer.
$endgroup$
– greedoid
4 hours ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
$begingroup$
You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying.
$endgroup$
– Matt Samuel
17 mins ago
add a comment |
$begingroup$
For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $vec b + c$ is not defined, and hence the expression $vec a cdot (vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.
Although it probably makes sense to you that
$$vec a cdot (vec b + c) = vec a cdot vec b + c cdot vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
$endgroup$
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
add a comment |
$begingroup$
For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $vec b + c$ is not defined, and hence the expression $vec a cdot (vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.
Although it probably makes sense to you that
$$vec a cdot (vec b + c) = vec a cdot vec b + c cdot vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
$endgroup$
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
add a comment |
$begingroup$
For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $vec b + c$ is not defined, and hence the expression $vec a cdot (vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.
Although it probably makes sense to you that
$$vec a cdot (vec b + c) = vec a cdot vec b + c cdot vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
$endgroup$
For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $vec b + c$ is not defined, and hence the expression $vec a cdot (vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.
Although it probably makes sense to you that
$$vec a cdot (vec b + c) = vec a cdot vec b + c cdot vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
edited 1 hour ago
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
answered 4 hours ago
Haris GusicHaris Gusic
1,142116
1,142116
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
add a comment |
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
The RHS expression isn't defined either.
$endgroup$
– Hammerite
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
@Hammerite You are right. What a silly mistake. Thank you!
$endgroup$
– Haris Gusic
2 hours ago
add a comment |
ItIsLastThursday is a new contributor. Be nice, and check out our Code of Conduct.
ItIsLastThursday is a new contributor. Be nice, and check out our Code of Conduct.
ItIsLastThursday is a new contributor. Be nice, and check out our Code of Conduct.
ItIsLastThursday is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3121781%2fdot-product-with-a-constant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This only makes sense if you make a vector space out of the base field
$endgroup$
– Brevan Ellefsen
1 hour ago