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Vector Algebra

Question
CBSEENMA12034261
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Let straight a with rightwards arrow on top space equals straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space space 7 space straight k with hat on top and straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top. Find a vector straight d with rightwards arrow on top which is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top comma space space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15.

Solution
straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space space 7 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top
   therefore space space 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 1 4 2 row 3 cell negative 2 end cell 7 end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell space space space 4 end cell 2 row cell negative 2 end cell 7 end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 2 row 3 7 end table close vertical bar plus space straight k with hat on top space open vertical bar table row 1 4 row 3 cell negative 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis 28 plus 4 right parenthesis space straight i with hat on top space minus space left parenthesis 7 minus 6 right parenthesis space straight j with hat on top plus space left parenthesis negative 2 minus 12 right parenthesis space straight k with hat on top space space space space space space space space space space space space space space space space space space space space space space equals space 32 space straight i with hat on top space minus space straight j with hat on top space minus space 14 space straight k with hat on top
which is a vector perpendicular to straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.
Since straight d with rightwards arrow on top is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.
therefore space space space space straight d with rightwards arrow on top space space is space parallel space to space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top. therefore space space space space space space space space space space space space space space straight d with rightwards arrow on top space equals space straight lambda left parenthesis straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top right parenthesis comma space space where space straight lambda space is space scalar space to space be space determined. therefore space space space space space space space space space space space space space space straight d with rightwards arrow on top space equals space straight lambda left parenthesis 32 straight i with hat on top space minus space straight j with hat on top space minus 14 space straight k with hat on top right parenthesis Now comma space space space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15 rightwards double arrow space space space space 2 space left parenthesis 32 space straight lambda right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis negative straight lambda right parenthesis space plus space left parenthesis 4 right parenthesis thin space left parenthesis negative 14 space straight lambda right parenthesis space equals space 15 rightwards double arrow space space space space 64 space straight lambda space plus space straight lambda space minus space 56 space straight lambda space equals space 15 space space space space space rightwards double arrow space space space 9 space straight lambda space equals space 15 space space space space rightwards double arrow space space space space straight lambda space equals space 15 over 9 space equals 5 over 3 therefore space space space space space space space space space space space space straight d with rightwards arrow on top space equals space 5 over 3 left parenthesis 32 space straight i with hat on top space minus space straight j with hat on top space minus space 14 space straight k with hat on top right parenthesis

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