Determine angle between 2 vectors.











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If there are two vectors given
$$vec {OA} = hat i + 2hat j~~text{and}~~vec {OB} = 4hat i + phat k $$



Then find the values of $p$ for which $angle AOB = cos^{-1}(frac{1}{5})$.



Would someone please give me a hand on solving this?






vectors angle







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    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    22 hours ago















up vote
-1
down vote

favorite












If there are two vectors given
$$vec {OA} = hat i + 2hat j~~text{and}~~vec {OB} = 4hat i + phat k $$



Then find the values of $p$ for which $angle AOB = cos^{-1}(frac{1}{5})$.



Would someone please give me a hand on solving this?






vectors angle







share|cite|improve this question









New contributor




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
















  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    22 hours ago













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











If there are two vectors given
$$vec {OA} = hat i + 2hat j~~text{and}~~vec {OB} = 4hat i + phat k $$



Then find the values of $p$ for which $angle AOB = cos^{-1}(frac{1}{5})$.



Would someone please give me a hand on solving this?






vectors angle







share|cite|improve this question









New contributor




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











If there are two vectors given
$$vec {OA} = hat i + 2hat j~~text{and}~~vec {OB} = 4hat i + phat k $$



Then find the values of $p$ for which $angle AOB = cos^{-1}(frac{1}{5})$.



Would someone please give me a hand on solving this?






vectors angle




vectors angle






share|cite|improve this question









New contributor




Mikail Khan 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









New contributor




Mikail Khan 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




share|cite|improve this question








edited 22 hours ago









Rakibul Islam Prince

80529




80529






New contributor




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









asked 23 hours ago









Mikail Khan

31




31




New contributor




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





New contributor





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






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








  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    22 hours ago














  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    22 hours ago








1




1




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
22 hours ago




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
22 hours ago










1 Answer
1






active

oldest

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up vote
0
down vote



accepted










HINT:



Try to use Dot product rule.which is
$$vec A. vec B=|vec A|.|vec B|.cos theta $$
Here,
$$vec A= hat i + 2hat j,vec B=4hat i + phat k$$
$$|vec A|=sqrt{1^2+2^2}=sqrt{5}~~,~~|vec B|=sqrt{16+p^2}$$
$$theta=cos^{-1}(frac{1}{5})$$



If you substitute everything in the formula I think you will get, $$p=pm 8$$






share|cite|improve this answer























  • Assalamualikum! it will be great if you show me substituted form, Thanks brother!
    – Mikail Khan
    22 hours ago










  • At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
    – Rakibul Islam Prince
    22 hours ago












  • Thank you so much man! I was stuck at the theta step, i forgot to put cos!
    – Mikail Khan
    22 hours ago










  • welcome to MSE family.
    – Rakibul Islam Prince
    22 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










HINT:



Try to use Dot product rule.which is
$$vec A. vec B=|vec A|.|vec B|.cos theta $$
Here,
$$vec A= hat i + 2hat j,vec B=4hat i + phat k$$
$$|vec A|=sqrt{1^2+2^2}=sqrt{5}~~,~~|vec B|=sqrt{16+p^2}$$
$$theta=cos^{-1}(frac{1}{5})$$



If you substitute everything in the formula I think you will get, $$p=pm 8$$






share|cite|improve this answer























  • Assalamualikum! it will be great if you show me substituted form, Thanks brother!
    – Mikail Khan
    22 hours ago










  • At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
    – Rakibul Islam Prince
    22 hours ago












  • Thank you so much man! I was stuck at the theta step, i forgot to put cos!
    – Mikail Khan
    22 hours ago










  • welcome to MSE family.
    – Rakibul Islam Prince
    22 hours ago















up vote
0
down vote



accepted










HINT:



Try to use Dot product rule.which is
$$vec A. vec B=|vec A|.|vec B|.cos theta $$
Here,
$$vec A= hat i + 2hat j,vec B=4hat i + phat k$$
$$|vec A|=sqrt{1^2+2^2}=sqrt{5}~~,~~|vec B|=sqrt{16+p^2}$$
$$theta=cos^{-1}(frac{1}{5})$$



If you substitute everything in the formula I think you will get, $$p=pm 8$$






share|cite|improve this answer























  • Assalamualikum! it will be great if you show me substituted form, Thanks brother!
    – Mikail Khan
    22 hours ago










  • At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
    – Rakibul Islam Prince
    22 hours ago












  • Thank you so much man! I was stuck at the theta step, i forgot to put cos!
    – Mikail Khan
    22 hours ago










  • welcome to MSE family.
    – Rakibul Islam Prince
    22 hours ago













up vote
0
down vote



accepted







up vote
0
down vote



accepted






HINT:



Try to use Dot product rule.which is
$$vec A. vec B=|vec A|.|vec B|.cos theta $$
Here,
$$vec A= hat i + 2hat j,vec B=4hat i + phat k$$
$$|vec A|=sqrt{1^2+2^2}=sqrt{5}~~,~~|vec B|=sqrt{16+p^2}$$
$$theta=cos^{-1}(frac{1}{5})$$



If you substitute everything in the formula I think you will get, $$p=pm 8$$






share|cite|improve this answer














HINT:



Try to use Dot product rule.which is
$$vec A. vec B=|vec A|.|vec B|.cos theta $$
Here,
$$vec A= hat i + 2hat j,vec B=4hat i + phat k$$
$$|vec A|=sqrt{1^2+2^2}=sqrt{5}~~,~~|vec B|=sqrt{16+p^2}$$
$$theta=cos^{-1}(frac{1}{5})$$



If you substitute everything in the formula I think you will get, $$p=pm 8$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 22 hours ago

























answered 22 hours ago









Rakibul Islam Prince

80529




80529












  • Assalamualikum! it will be great if you show me substituted form, Thanks brother!
    – Mikail Khan
    22 hours ago










  • At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
    – Rakibul Islam Prince
    22 hours ago












  • Thank you so much man! I was stuck at the theta step, i forgot to put cos!
    – Mikail Khan
    22 hours ago










  • welcome to MSE family.
    – Rakibul Islam Prince
    22 hours ago


















  • Assalamualikum! it will be great if you show me substituted form, Thanks brother!
    – Mikail Khan
    22 hours ago










  • At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
    – Rakibul Islam Prince
    22 hours ago












  • Thank you so much man! I was stuck at the theta step, i forgot to put cos!
    – Mikail Khan
    22 hours ago










  • welcome to MSE family.
    – Rakibul Islam Prince
    22 hours ago
















Assalamualikum! it will be great if you show me substituted form, Thanks brother!
– Mikail Khan
22 hours ago




Assalamualikum! it will be great if you show me substituted form, Thanks brother!
– Mikail Khan
22 hours ago












At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
– Rakibul Islam Prince
22 hours ago






At last you should get something like $4=sqrt{5}.sqrt{16+p^2}.cosbigg(cos^{-1}(frac{1}{5})bigg)$
– Rakibul Islam Prince
22 hours ago














Thank you so much man! I was stuck at the theta step, i forgot to put cos!
– Mikail Khan
22 hours ago




Thank you so much man! I was stuck at the theta step, i forgot to put cos!
– Mikail Khan
22 hours ago












welcome to MSE family.
– Rakibul Islam Prince
22 hours ago




welcome to MSE family.
– Rakibul Islam Prince
22 hours ago










Mikail Khan is a new contributor. Be nice, and check out our Code of Conduct.










 

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Mikail Khan is a new contributor. Be nice, and check out our Code of Conduct.













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