If <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#6 {main}</pre> then do you think that A is necessarily equal to C?


Given that,

                         

Let angle between  be  and that between 

∴     

      

                

Now,    if   then A = C, and

              then 

Thus A is not necessarily equal to C.

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Find the unit vector along the resultant of left parenthesis straight i with hat on top plus straight j with hat on top right parenthesis space and space left parenthesis straight i with hat on top minus straight j with hat on top right parenthesis.


Resultant of  is

 
∴   Unit vector in direction of 

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Find a vector having magnitude equal to the magnitude of vector straight A with rightwards arrow on top space equals space minus 2 straight i with hat on top plus 1 straight j with hat on top plus 4 straight k with hat on top and parallel to vector straight B with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top minus 3 stack straight k. with hat on top


Magnitude of vector A is,

     

Unit vector in direction of vector 



The vector that has magnitude same as that of vector and parallel to vector 



      

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If straight A with rightwards arrow on top times straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top times straight C with rightwards arrow on top and angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top is twice the angle between <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> then show that straight C over straight B equals 2 cos straight theta over 2 minus sec straight theta over 2 comma where straight theta is the angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top.


It is given that angle between  is  

Therefore the angle between  will be 

Here,        

∴         

               


               .

Hence proved. 

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Prove that:
left parenthesis 3 straight A with rightwards arrow on top plus 2 straight B with rightwards arrow on top right parenthesis times left parenthesis 2 straight A with rightwards arrow on top minus 3 straight B with rightwards arrow on top right parenthesis space equals space 6 straight A squared minus 6 straight B squared minus 5 ABcosθ, where straight theta is the angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top.


By using distributive and commutative law, the dot product can be evaluated as, 

   

     
 


                  



That is,

     

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