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Vector Algebra

Question
CBSEENMA12034021
Wired Faculty App

The vectors of magnitude a, 2a, 3a meet at a point and their directions arc. along the diagonals of three adjacent faces of a cube. Determine the magnitude of their resultant.

Solution

Let straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top be unit vectors along OA, OC, OE respectively. Let OB, OD, OF be the diagonals of faces OABC, OCDE, OEFA respectively of the cube of side of length b.

  Now  OB with rightwards arrow on top space equals OA with rightwards arrow on top space plus space AB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space OC with rightwards arrow on top space equals space straight b space straight i with hat on top space plus space straight b space straight j with hat on top
therefore space space space open vertical bar OB with rightwards arrow on top close vertical bar space equals space square root of straight b squared plus straight b squared end root space equals space square root of 2 space straight b squared end root space equals space square root of 2 space straight b
     therefore   unit vector along
                  OB with rightwards arrow on top space equals space fraction numerator OB with rightwards arrow on top over denominator open vertical bar OB with rightwards arrow on top close vertical bar end fraction equals space fraction numerator straight b space straight i with hat on top space plus space straight b space straight j with hat on top over denominator square root of 2 space straight b end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis
      Similarly unit vector along OD with rightwards arrow on top = fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight j with hat on top plus straight k with hat on top right parenthesis
  and unit vector along OF with rightwards arrow on top space equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis
       therefore space space space space space OB with rightwards arrow on top space equals space fraction numerator straight a with rightwards arrow on top over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis comma space space space space OD with rightwards arrow on top space equals space fraction numerator 2 straight a over denominator square root of 2 end fraction left parenthesis straight j with hat on top space plus straight k with hat on top right parenthesis comma space space OF with rightwards arrow on top space equals space fraction numerator 3 straight a over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis
Let straight R with rightwards arrow on top be resultant of vectors OB with rightwards arrow on top comma space OD with rightwards arrow on top comma space OF with rightwards arrow on top
therefore space space space space space space space straight R with rightwards arrow on top space equals space OB with rightwards arrow on top space plus space OD with rightwards arrow on top space plus space OF with rightwards arrow on top space space space space space space space space space space space space space space space space equals fraction numerator straight a over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis space fraction numerator 2 straight a over denominator square root of 2 end fraction left parenthesis straight j with hat on top space plus space straight k with hat on top right parenthesis space plus space fraction numerator 3 straight a over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis space space space space space space space space space space space space space space space space equals space fraction numerator 4 straight a over denominator square root of 2 end fraction straight i with hat on top space plus space fraction numerator 3 straight a over denominator square root of 2 end fraction straight j with hat on top space plus space fraction numerator 5 straight a over denominator square root of 2 end fraction straight k with hat on top
therefore      magnitude of resultant vector =  open vertical bar straight R with rightwards arrow on top close vertical bar
                              equals space square root of open parentheses fraction numerator 4 straight a over denominator square root of 2 end fraction close parentheses squared plus open parentheses fraction numerator 3 space straight a over denominator square root of 2 end fraction close parentheses squared plus space open parentheses fraction numerator 5 straight a over denominator square root of 2 end fraction close parentheses squared end root equals space square root of fraction numerator 16 straight a squared over denominator 2 end fraction plus fraction numerator 9 straight a squared over denominator 2 end fraction plus fraction numerator 25 space straight a squared over denominator 2 end fraction end root space equals space square root of 25 space straight a squared end root space equals space 5 space straight a
                      
                     
 

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