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Question - Vectors [on hold]

-3

$begingroup$

$left|uright| = 1, left|vright|=2$
$left|3u-5vright|^2 + left|2u-3vright|^2 = 107$
how can i get the angle between the two vectors?

vectors

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edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

put on hold as off-topic by John Douma, Michael Hoppe, Leucippus, Shailesh, Alex Provost Apr 9 at 4:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, Michael Hoppe, Leucippus, Shailesh, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

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  – J. W. Tanner
  Apr 8 at 13:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – Alex Provost
  Apr 9 at 4:15
  

  
  
  
  

add a comment | 

-3

$begingroup$

$left|uright| = 1, left|vright|=2$
$left|3u-5vright|^2 + left|2u-3vright|^2 = 107$
how can i get the angle between the two vectors?

vectors

share|cite|improve this question

edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by John Douma, Michael Hoppe, Leucippus, Shailesh, Alex Provost Apr 9 at 4:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, Michael Hoppe, Leucippus, Shailesh, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Please use MathJax
  $endgroup$
  – J. W. Tanner
  Apr 8 at 13:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – Alex Provost
  Apr 9 at 4:15
  

  
  
  
  

add a comment | 

$begingroup$

$left|uright| = 1, left|vright|=2$
$left|3u-5vright|^2 + left|2u-3vright|^2 = 107$
how can i get the angle between the two vectors?

vectors

share|cite|improve this question

edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 8 at 14:12

Don Thousand

4,569734

edited Apr 8 at 14:12

Don Thousand

4,569734

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

New contributor

iSkullmao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 13:09

iSkullmaoiSkullmao

11

2 Answers
2

active

oldest

votes

0

$begingroup$

Let $u=(a,b),v=(c,d)$. So, $a^2+b^2=1,;c^2+d^2=4$. And finally, $$(3a-5c)^2+(3b-5d)^2+(2a-3c)^2+(2b-3d)^2=107$$$$13(a^2+b^2)-42(ac+bd)+34(c^2+d^2)=107$$$$13+136-42(ac+bd)=107$$$$-42(vcdot u)=-42$$$$|v||u|sintheta=1$$$$sintheta=frac12$$

share|cite|improve this answer

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your effort, but this isn't correct answer. I have just found the answer using the fact that |3u-5v|^2 is equal to the dot product of (3u-5v) . (3u-5v) . then i got the value of u.v then used the rule : cos x = (a.b)/( |a|.|b|) and the answer was pi*3
  $endgroup$
  – iSkullmao
  Apr 8 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iSkullmao Uhhh, what do you think $sin(fracpi3)$ is lolll.
  $endgroup$
  – Don Thousand
  Apr 8 at 13:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  pretty sure i saw |v||u| sinx = 1/21 sinx = 1/42 youre so sneaky😂
  $endgroup$
  – iSkullmao
  Apr 8 at 13:44
  

  
  
  
  

add a comment | 

0

$begingroup$

You can use the formula $costheta = fracucdot v$.

$|u|$ and $|v|$ are already given, so you know the denominator is $2$.

Now let u = (a, b) and v = (c, d).

Applying that to the next equation, we have $$(3a - 5c)^2 + (3b - 5d)^2 + (2a - 3c)^2 + (2b - 3d)^2 = 107$$

Now expanding the terms and combining like variables:
$$9a^2 - 15ac + 25c^2 + 4a^2 - 6ac + 9c^2 = 13a^2 - 21ac + 34c^2$$
and
$$9b^2 - 15bd + 25d^2 + 4b^2 - 6bd + 9d^2 = 13b^2 - 21bd + 34d^2$$

Then factoring out the constants you have
$$13(a^2 + b^2) + 34(c^2 + d^2) - 21(ac + bd) = 107$$

Note that the first two terms contain $|u|^2$ and $|v|^2$, and the third term is the dot product $ucdot v$.

Then we have $13 + 136 - 21ucdot v = 107$, which simplifies to $42 = 21ucdot v$, and
$ucdot v = 2$.

Substituting into the angle formula, $costheta = 2/2 = 1$, thus $theta = 0$.

share|cite|improve this answer

answered Apr 8 at 14:16

VahanVahan

849

$endgroup$

add a comment | 

0

$begingroup$

Let $u=(a,b),v=(c,d)$. So, $a^2+b^2=1,;c^2+d^2=4$. And finally, $$(3a-5c)^2+(3b-5d)^2+(2a-3c)^2+(2b-3d)^2=107$$$$13(a^2+b^2)-42(ac+bd)+34(c^2+d^2)=107$$$$13+136-42(ac+bd)=107$$$$-42(vcdot u)=-42$$$$|v||u|sintheta=1$$$$sintheta=frac12$$

share|cite|improve this answer

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your effort, but this isn't correct answer. I have just found the answer using the fact that |3u-5v|^2 is equal to the dot product of (3u-5v) . (3u-5v) . then i got the value of u.v then used the rule : cos x = (a.b)/( |a|.|b|) and the answer was pi*3
  $endgroup$
  – iSkullmao
  Apr 8 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iSkullmao Uhhh, what do you think $sin(fracpi3)$ is lolll.
  $endgroup$
  – Don Thousand
  Apr 8 at 13:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  pretty sure i saw |v||u| sinx = 1/21 sinx = 1/42 youre so sneaky😂
  $endgroup$
  – iSkullmao
  Apr 8 at 13:44
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $u=(a,b),v=(c,d)$. So, $a^2+b^2=1,;c^2+d^2=4$. And finally, $$(3a-5c)^2+(3b-5d)^2+(2a-3c)^2+(2b-3d)^2=107$$$$13(a^2+b^2)-42(ac+bd)+34(c^2+d^2)=107$$$$13+136-42(ac+bd)=107$$$$-42(vcdot u)=-42$$$$|v||u|sintheta=1$$$$sintheta=frac12$$

share|cite|improve this answer

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your effort, but this isn't correct answer. I have just found the answer using the fact that |3u-5v|^2 is equal to the dot product of (3u-5v) . (3u-5v) . then i got the value of u.v then used the rule : cos x = (a.b)/( |a|.|b|) and the answer was pi*3
  $endgroup$
  – iSkullmao
  Apr 8 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @iSkullmao Uhhh, what do you think $sin(fracpi3)$ is lolll.
  $endgroup$
  – Don Thousand
  Apr 8 at 13:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  pretty sure i saw |v||u| sinx = 1/21 sinx = 1/42 youre so sneaky😂
  $endgroup$
  – iSkullmao
  Apr 8 at 13:44
  

  
  
  
  

add a comment | 

$begingroup$

Let $u=(a,b),v=(c,d)$. So, $a^2+b^2=1,;c^2+d^2=4$. And finally, $$(3a-5c)^2+(3b-5d)^2+(2a-3c)^2+(2b-3d)^2=107$$$$13(a^2+b^2)-42(ac+bd)+34(c^2+d^2)=107$$$$13+136-42(ac+bd)=107$$$$-42(vcdot u)=-42$$$$|v||u|sintheta=1$$$$sintheta=frac12$$

share|cite|improve this answer

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

answered Apr 8 at 13:25

Don ThousandDon Thousand

4,569734

0

$begingroup$

You can use the formula $costheta = fracucdot v$.

$|u|$ and $|v|$ are already given, so you know the denominator is $2$.

Now let u = (a, b) and v = (c, d).

Applying that to the next equation, we have $$(3a - 5c)^2 + (3b - 5d)^2 + (2a - 3c)^2 + (2b - 3d)^2 = 107$$

Now expanding the terms and combining like variables:
$$9a^2 - 15ac + 25c^2 + 4a^2 - 6ac + 9c^2 = 13a^2 - 21ac + 34c^2$$
and
$$9b^2 - 15bd + 25d^2 + 4b^2 - 6bd + 9d^2 = 13b^2 - 21bd + 34d^2$$

Then factoring out the constants you have
$$13(a^2 + b^2) + 34(c^2 + d^2) - 21(ac + bd) = 107$$

Note that the first two terms contain $|u|^2$ and $|v|^2$, and the third term is the dot product $ucdot v$.

Then we have $13 + 136 - 21ucdot v = 107$, which simplifies to $42 = 21ucdot v$, and
$ucdot v = 2$.

Substituting into the angle formula, $costheta = 2/2 = 1$, thus $theta = 0$.

share|cite|improve this answer

answered Apr 8 at 14:16

VahanVahan

849

$endgroup$

add a comment | 

0

$begingroup$

You can use the formula $costheta = fracucdot v$.

$|u|$ and $|v|$ are already given, so you know the denominator is $2$.

Now let u = (a, b) and v = (c, d).

Applying that to the next equation, we have $$(3a - 5c)^2 + (3b - 5d)^2 + (2a - 3c)^2 + (2b - 3d)^2 = 107$$

Now expanding the terms and combining like variables:
$$9a^2 - 15ac + 25c^2 + 4a^2 - 6ac + 9c^2 = 13a^2 - 21ac + 34c^2$$
and
$$9b^2 - 15bd + 25d^2 + 4b^2 - 6bd + 9d^2 = 13b^2 - 21bd + 34d^2$$

Then factoring out the constants you have
$$13(a^2 + b^2) + 34(c^2 + d^2) - 21(ac + bd) = 107$$

Note that the first two terms contain $|u|^2$ and $|v|^2$, and the third term is the dot product $ucdot v$.

Then we have $13 + 136 - 21ucdot v = 107$, which simplifies to $42 = 21ucdot v$, and
$ucdot v = 2$.

Substituting into the angle formula, $costheta = 2/2 = 1$, thus $theta = 0$.

share|cite|improve this answer

answered Apr 8 at 14:16

VahanVahan

849

$endgroup$

add a comment | 

$begingroup$

You can use the formula $costheta = fracucdot v$.

$|u|$ and $|v|$ are already given, so you know the denominator is $2$.

Now let u = (a, b) and v = (c, d).

Applying that to the next equation, we have $$(3a - 5c)^2 + (3b - 5d)^2 + (2a - 3c)^2 + (2b - 3d)^2 = 107$$

Now expanding the terms and combining like variables:
$$9a^2 - 15ac + 25c^2 + 4a^2 - 6ac + 9c^2 = 13a^2 - 21ac + 34c^2$$
and
$$9b^2 - 15bd + 25d^2 + 4b^2 - 6bd + 9d^2 = 13b^2 - 21bd + 34d^2$$

Then factoring out the constants you have
$$13(a^2 + b^2) + 34(c^2 + d^2) - 21(ac + bd) = 107$$

Note that the first two terms contain $|u|^2$ and $|v|^2$, and the third term is the dot product $ucdot v$.

Then we have $13 + 136 - 21ucdot v = 107$, which simplifies to $42 = 21ucdot v$, and
$ucdot v = 2$.

Substituting into the angle formula, $costheta = 2/2 = 1$, thus $theta = 0$.

share|cite|improve this answer

answered Apr 8 at 14:16

VahanVahan

849

$endgroup$

share|cite|improve this answer

answered Apr 8 at 14:16

VahanVahan

849

answered Apr 8 at 14:16

VahanVahan

849

answered Apr 8 at 14:16

VahanVahan

849