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Vector Algebra

Question
CBSEENMA12034166
Wired Faculty App

Find open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar comma space space space if space space straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space plus straight j with hat on top space plus space 3 straight k with hat on top space and space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 2 space straight k with hat on top.

Solution

Here,    straight a with rightwards arrow on top space equals space 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 2 space straight k with hat on top
therefore space space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 1 3 row 3 5 cell negative 2 end cell end table close vertical bar space equals space straight i with hat on top space open vertical bar table row 1 3 row 5 cell negative 2 end cell end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 2 3 row 3 cell negative 2 end cell end table close vertical bar plus space straight k with hat on top space open vertical bar table row 2 1 row 3 5 end table close vertical bar
                     equals space left parenthesis negative 2 minus 15 right parenthesis space straight i with hat on top space minus space left parenthesis negative 4 minus 9 right parenthesis space straight j with hat on top space plus space left parenthesis 10 minus 3 right parenthesis straight k with hat on top space equals space minus 17 straight i with hat on top space plus space 13 space straight j with hat on top space plus space 7 space straight k with hat on top
therefore space space space open vertical bar straight a with rightwards arrow on top cross times space straight b with rightwards arrow on top close vertical bar space equals space square root of left parenthesis negative 17 right parenthesis squared plus left parenthesis 13 right parenthesis squared plus left parenthesis 7 right parenthesis squared end root space equals space 289 plus 169 plus 49 space equals space square root of 507
                 

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