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Vector Algebra

Question
CBSEENMA12034134
Wired Faculty App

If straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top are mutually perpendicular vectors of equal magnitude, show that they are equally inclined to the vector left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis

Solution

We have
                 open vertical bar straight a with rightwards arrow on top close vertical bar space equals space open vertical bar straight b with rightwards arrow on top close vertical bar space equals open vertical bar straight c with rightwards arrow on top close vertical bar space and space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space straight b with rightwards arrow on top. space straight c with rightwards arrow on top space equals space straight c with rightwards arrow on top. space straight a with rightwards arrow on top space equals space 0                 ...(1)
 Now space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar squared space equals space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis squared space equals space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis. space left parenthesis straight a with rightwards arrow on top plus space straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space space space space space space space space space space space space space space space space space space space space space space equals space straight a with rightwards arrow on top. space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top. space straight b with rightwards arrow on top space plus space straight c with rightwards arrow on top. space straight c with rightwards arrow on top space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus open vertical bar straight c with rightwards arrow on top close vertical bar squared space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket space space space space space space space space space space space space space space space space space space space space space space space equals open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals space 3 space open vertical bar straight a with rightwards arrow on top close vertical bar squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket therefore space space space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space equals space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Let straight theta be angle between <pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...') #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...') #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #6 /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) #7 {main}</pre>
therefore space space straight a with rightwards arrow on top left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space equals space open vertical bar straight a with rightwards arrow on top close vertical bar space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space cos space straight theta therefore space space space space space space space space space straight a with rightwards arrow on top. space straight a with rightwards arrow on top space equals space open vertical bar straight a with rightwards arrow on top close vertical bar. space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar space cos space straight theta space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets rightwards double arrow space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar squared space cos space straight theta space space space space space rightwards double arrow space space space cos space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space space space space space straight theta space equals space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses.
Let straight theta be the angle between
 straight b with rightwards arrow on top space and space straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top
  therefore space space space straight b with rightwards arrow on top space. space space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space equals space open vertical bar straight b with rightwards arrow on top close vertical bar space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space cos space straight ϕ therefore space space space space straight b with rightwards arrow on top. space straight b with rightwards arrow on top space equals space open vertical bar straight b with rightwards arrow on top close vertical bar space. space square root of 3 space open vertical bar straight b with rightwards arrow on top close vertical bar space cos space straight ϕ space space space space space space space left square bracket because space space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space equals space square root of 3 space open vertical bar straight b with rightwards arrow on top close vertical bar right square bracket rightwards double arrow space space space space open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space square root of 3 space open vertical bar straight b with rightwards arrow on top close vertical bar squared space cos space straight ϕ space space space space rightwards double arrow space space space space cos space straight ϕ space equals space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space space space space straight ϕ space equals space cos space to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
Let straight psi be angle between straight c with rightwards arrow on top space and space straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top
therefore space space space space straight c with rightwards arrow on top. space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space equals space open vertical bar straight c with rightwards arrow on top close vertical bar space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space cos space straight psi
therefore space space straight c with rightwards arrow on top. space straight c with rightwards arrow on top space equals open vertical bar straight c with rightwards arrow on top close vertical bar. space square root of 3 space open vertical bar straight c with rightwards arrow on top close vertical bar space cos space straight psi rightwards double arrow space space space space space space space space straight psi space equals space cos to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses therefore space space space space space space space straight theta space equals straight ϕ space equals straight psi therefore space space space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top space are space equally space inclined space to space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis.

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