Determine vector c, which is collinear vector of vector $a+b$
$begingroup$
Determine vector c, which is collinear vector of vector $a+b$, if $ab=5$, $cb=18$ and $|b|=2$.
I tried with $c=n(a+b)$.
$9= |c|*cos(alpha)$...$|c|=sqrt{(n^2(a+b)(a+b))}= n sqrt{aa+14}$
Then $9= n sqrt{aa+14}*cos(alpha)$
Second equation is: $frac{5}{2} sqrt{aa}*cos(alpha)$
From second equation we get: $cos(alpha)=frac{5}{2sqrt{aa}}$. I put this in first equation and I get that $n=frac{9sqrt{14}}{35}$
My solution is: $c=frac{9sqrt{14}}{35}(a+b)$
Is this correct?
Thank you for your help.
vectors inner-product-space
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
$endgroup$
add a comment |
$begingroup$
Determine vector c, which is collinear vector of vector $a+b$, if $ab=5$, $cb=18$ and $|b|=2$.
I tried with $c=n(a+b)$.
$9= |c|*cos(alpha)$...$|c|=sqrt{(n^2(a+b)(a+b))}= n sqrt{aa+14}$
Then $9= n sqrt{aa+14}*cos(alpha)$
Second equation is: $frac{5}{2} sqrt{aa}*cos(alpha)$
From second equation we get: $cos(alpha)=frac{5}{2sqrt{aa}}$. I put this in first equation and I get that $n=frac{9sqrt{14}}{35}$
My solution is: $c=frac{9sqrt{14}}{35}(a+b)$
Is this correct?
Thank you for your help.
vectors inner-product-space
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
$endgroup$
add a comment |
$begingroup$
Determine vector c, which is collinear vector of vector $a+b$, if $ab=5$, $cb=18$ and $|b|=2$.
I tried with $c=n(a+b)$.
$9= |c|*cos(alpha)$...$|c|=sqrt{(n^2(a+b)(a+b))}= n sqrt{aa+14}$
Then $9= n sqrt{aa+14}*cos(alpha)$
Second equation is: $frac{5}{2} sqrt{aa}*cos(alpha)$
From second equation we get: $cos(alpha)=frac{5}{2sqrt{aa}}$. I put this in first equation and I get that $n=frac{9sqrt{14}}{35}$
My solution is: $c=frac{9sqrt{14}}{35}(a+b)$
Is this correct?
Thank you for your help.
vectors inner-product-space
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
$endgroup$
Determine vector c, which is collinear vector of vector $a+b$, if $ab=5$, $cb=18$ and $|b|=2$.
I tried with $c=n(a+b)$.
$9= |c|*cos(alpha)$...$|c|=sqrt{(n^2(a+b)(a+b))}= n sqrt{aa+14}$
Then $9= n sqrt{aa+14}*cos(alpha)$
Second equation is: $frac{5}{2} sqrt{aa}*cos(alpha)$
From second equation we get: $cos(alpha)=frac{5}{2sqrt{aa}}$. I put this in first equation and I get that $n=frac{9sqrt{14}}{35}$
My solution is: $c=frac{9sqrt{14}}{35}(a+b)$
Is this correct?
Thank you for your help.
vectors inner-product-space
vectors inner-product-space
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
edited Jan 6 at 13:15
Michael Rozenberg
108k1895200
108k1895200
asked Jan 6 at 11:43
J.DoeJ.Doe
899
asked Jan 6 at 11:43
J.DoeJ.Doe
899
asked Jan 6 at 11:43
J.DoeJ.Doe
899
899
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$n(a+b)b=18$$ or
$$n(5+4)=18$$ or
$$n=2,$$ which gives
$$c=2(a+b).$$
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
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
});
}
});
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%2f3063754%2fdetermine-vector-c-which-is-collinear-vector-of-vector-ab%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$n(a+b)b=18$$ or
$$n(5+4)=18$$ or
$$n=2,$$ which gives
$$c=2(a+b).$$
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$n(a+b)b=18$$ or
$$n(5+4)=18$$ or
$$n=2,$$ which gives
$$c=2(a+b).$$
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$$n(a+b)b=18$$ or
$$n(5+4)=18$$ or
$$n=2,$$ which gives
$$c=2(a+b).$$
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$$n(a+b)b=18$$ or
$$n(5+4)=18$$ or
$$n=2,$$ which gives
$$c=2(a+b).$$
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 6 at 13:11
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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%2f3063754%2fdetermine-vector-c-which-is-collinear-vector-of-vector-ab%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
