Vector dot products












0












$begingroup$



Consider two vectors $vec{a:}$, $vec{b:}$ with $|vec{a:}| = 2,
> |vec{b:}| = 1$, and an angle between them of $60^circ$ . Find the scalar
product$$left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)$$




So I know that $left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)= left|vec{a:}right|^2-left|vec{b:}right|^2$



But in the solution it says that it is equal to $$left|vec{a:}right|^2-left|vec{a:}right|left|vec{b:}right|cos:60-2left|vec{b:}right|^2$$



Why though?



edit: Here are the pics
Question



Solution






calculus vector-spaces vectors







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  • $begingroup$
    Your answer is correct
    $endgroup$
    – Ankit Kumar
    Dec 23 '18 at 16:45






  • 8




    $begingroup$
    Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 16:45










  • $begingroup$
    No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
    $endgroup$
    – Naochi
    Dec 23 '18 at 17:05






  • 1




    $begingroup$
    @Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:24










  • $begingroup$
    When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:29
















0












$begingroup$



Consider two vectors $vec{a:}$, $vec{b:}$ with $|vec{a:}| = 2,
> |vec{b:}| = 1$, and an angle between them of $60^circ$ . Find the scalar
product$$left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)$$




So I know that $left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)= left|vec{a:}right|^2-left|vec{b:}right|^2$



But in the solution it says that it is equal to $$left|vec{a:}right|^2-left|vec{a:}right|left|vec{b:}right|cos:60-2left|vec{b:}right|^2$$



Why though?



edit: Here are the pics
Question



Solution






calculus vector-spaces vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your answer is correct
    $endgroup$
    – Ankit Kumar
    Dec 23 '18 at 16:45






  • 8




    $begingroup$
    Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 16:45










  • $begingroup$
    No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
    $endgroup$
    – Naochi
    Dec 23 '18 at 17:05






  • 1




    $begingroup$
    @Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:24










  • $begingroup$
    When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:29














0












0








0





$begingroup$



Consider two vectors $vec{a:}$, $vec{b:}$ with $|vec{a:}| = 2,
> |vec{b:}| = 1$, and an angle between them of $60^circ$ . Find the scalar
product$$left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)$$




So I know that $left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)= left|vec{a:}right|^2-left|vec{b:}right|^2$



But in the solution it says that it is equal to $$left|vec{a:}right|^2-left|vec{a:}right|left|vec{b:}right|cos:60-2left|vec{b:}right|^2$$



Why though?



edit: Here are the pics
Question



Solution






calculus vector-spaces vectors







share|cite|improve this question











$endgroup$





Consider two vectors $vec{a:}$, $vec{b:}$ with $|vec{a:}| = 2,
> |vec{b:}| = 1$, and an angle between them of $60^circ$ . Find the scalar
product$$left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)$$




So I know that $left(vec{a:}+vec{b:}right)cdot left(vec{a:}-vec{b:}right)= left|vec{a:}right|^2-left|vec{b:}right|^2$



But in the solution it says that it is equal to $$left|vec{a:}right|^2-left|vec{a:}right|left|vec{b:}right|cos:60-2left|vec{b:}right|^2$$



Why though?



edit: Here are the pics
Question



Solution






calculus vector-spaces vectors




calculus vector-spaces vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 1:06









lioness99a

3,8362727




3,8362727










asked Dec 23 '18 at 16:40









NaochiNaochi

626




626












  • $begingroup$
    Your answer is correct
    $endgroup$
    – Ankit Kumar
    Dec 23 '18 at 16:45






  • 8




    $begingroup$
    Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 16:45










  • $begingroup$
    No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
    $endgroup$
    – Naochi
    Dec 23 '18 at 17:05






  • 1




    $begingroup$
    @Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:24










  • $begingroup$
    When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:29


















  • $begingroup$
    Your answer is correct
    $endgroup$
    – Ankit Kumar
    Dec 23 '18 at 16:45






  • 8




    $begingroup$
    Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 16:45










  • $begingroup$
    No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
    $endgroup$
    – Naochi
    Dec 23 '18 at 17:05






  • 1




    $begingroup$
    @Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:24










  • $begingroup$
    When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:29
















$begingroup$
Your answer is correct
$endgroup$
– Ankit Kumar
Dec 23 '18 at 16:45




$begingroup$
Your answer is correct
$endgroup$
– Ankit Kumar
Dec 23 '18 at 16:45




8




8




$begingroup$
Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
$endgroup$
– Mark Heavey
Dec 23 '18 at 16:45




$begingroup$
Are you sure it's not a typo? The bottom line there is equal to $ ( textbf{a} - 2 textbf{b} ) cdot ( textbf{a} + textbf{b} )$.
$endgroup$
– Mark Heavey
Dec 23 '18 at 16:45












$begingroup$
No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
$endgroup$
– Naochi
Dec 23 '18 at 17:05




$begingroup$
No it is correct, the professor confirmed it, but I missed the explanation of why it is like this so I was hoping to get some answers here. I will post a screenshot of it.
$endgroup$
– Naochi
Dec 23 '18 at 17:05




1




1




$begingroup$
@Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
$endgroup$
– Eric Towers
Dec 23 '18 at 17:24




$begingroup$
@Naochi : Your large quotation has "$(vec{a}+vec{b}) cdot (vec{a} - vec{b})$". Your pic has $(vec{a}+vec{b}) cdot (vec{a} - 2vec{b})$". Which one do you want to talk about?
$endgroup$
– Eric Towers
Dec 23 '18 at 17:24












$begingroup$
When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
$endgroup$
– Eric Towers
Dec 23 '18 at 17:29




$begingroup$
When you say "So I know that ...", are you thinking this is the difference of two squares factorization?
$endgroup$
– Eric Towers
Dec 23 '18 at 17:29










2 Answers
2






active

oldest

votes


















0












$begingroup$

the expression is given
$$(vec a + vec b).(vec a - vec b)=vec a.vec a-vec a.vec b+vec b.vec a-vec b.vec b$$
$$=|vec a|.|vec a|cos 0^o-|vec a|.|vec b|cos 60^o+|vec a|.|vec b|cos 60^o-|vec b|.|vec b|cos 0^o$$
$$=|vec a|.|vec a|-|vec b|.|vec b|$$
$$=|vec a|^2-|vec b|^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
    $endgroup$
    – Rakibul Islam Prince
    Dec 23 '18 at 23:00





















0












$begingroup$

Note that you ask a question about
$$
(a+b) (a-b)
$$
but the question in the book is about
$$
(a+b) (a-2b)
$$
I think you missed the factor of $2$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    the expression is given
    $$(vec a + vec b).(vec a - vec b)=vec a.vec a-vec a.vec b+vec b.vec a-vec b.vec b$$
    $$=|vec a|.|vec a|cos 0^o-|vec a|.|vec b|cos 60^o+|vec a|.|vec b|cos 60^o-|vec b|.|vec b|cos 0^o$$
    $$=|vec a|.|vec a|-|vec b|.|vec b|$$
    $$=|vec a|^2-|vec b|^2$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
      $endgroup$
      – Rakibul Islam Prince
      Dec 23 '18 at 23:00


















    0












    $begingroup$

    the expression is given
    $$(vec a + vec b).(vec a - vec b)=vec a.vec a-vec a.vec b+vec b.vec a-vec b.vec b$$
    $$=|vec a|.|vec a|cos 0^o-|vec a|.|vec b|cos 60^o+|vec a|.|vec b|cos 60^o-|vec b|.|vec b|cos 0^o$$
    $$=|vec a|.|vec a|-|vec b|.|vec b|$$
    $$=|vec a|^2-|vec b|^2$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
      $endgroup$
      – Rakibul Islam Prince
      Dec 23 '18 at 23:00
















    0












    0








    0





    $begingroup$

    the expression is given
    $$(vec a + vec b).(vec a - vec b)=vec a.vec a-vec a.vec b+vec b.vec a-vec b.vec b$$
    $$=|vec a|.|vec a|cos 0^o-|vec a|.|vec b|cos 60^o+|vec a|.|vec b|cos 60^o-|vec b|.|vec b|cos 0^o$$
    $$=|vec a|.|vec a|-|vec b|.|vec b|$$
    $$=|vec a|^2-|vec b|^2$$






    share|cite|improve this answer











    $endgroup$



    the expression is given
    $$(vec a + vec b).(vec a - vec b)=vec a.vec a-vec a.vec b+vec b.vec a-vec b.vec b$$
    $$=|vec a|.|vec a|cos 0^o-|vec a|.|vec b|cos 60^o+|vec a|.|vec b|cos 60^o-|vec b|.|vec b|cos 0^o$$
    $$=|vec a|.|vec a|-|vec b|.|vec b|$$
    $$=|vec a|^2-|vec b|^2$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 22:59

























    answered Dec 23 '18 at 17:49









    Rakibul Islam PrinceRakibul Islam Prince

    988211




    988211












    • $begingroup$
      sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
      $endgroup$
      – Rakibul Islam Prince
      Dec 23 '18 at 23:00




















    • $begingroup$
      sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
      $endgroup$
      – Rakibul Islam Prince
      Dec 23 '18 at 23:00


















    $begingroup$
    sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
    $endgroup$
    – Rakibul Islam Prince
    Dec 23 '18 at 23:00






    $begingroup$
    sorry I meant to say their magnitude but had written in their vector notation.Now it is fixed.thanks ...
    $endgroup$
    – Rakibul Islam Prince
    Dec 23 '18 at 23:00













    0












    $begingroup$

    Note that you ask a question about
    $$
    (a+b) (a-b)
    $$
    but the question in the book is about
    $$
    (a+b) (a-2b)
    $$
    I think you missed the factor of $2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Note that you ask a question about
      $$
      (a+b) (a-b)
      $$
      but the question in the book is about
      $$
      (a+b) (a-2b)
      $$
      I think you missed the factor of $2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Note that you ask a question about
        $$
        (a+b) (a-b)
        $$
        but the question in the book is about
        $$
        (a+b) (a-2b)
        $$
        I think you missed the factor of $2$.






        share|cite|improve this answer









        $endgroup$



        Note that you ask a question about
        $$
        (a+b) (a-b)
        $$
        but the question in the book is about
        $$
        (a+b) (a-2b)
        $$
        I think you missed the factor of $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 22:59









        EddyEddy

        959612




        959612






























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