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Vector Algebra

Question
CBSEENMA12034186
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Find the value of λ so that the two vectors 2 straight i with hat on top space plus space 3 straight j with hat on top space minus straight k with hat on top and 4 straight i with hat on top plus 6 straight j with hat on top plus straight lambda straight k with hat on top are (i) parallel (ii) perpendicular to each other. 

Solution

Let straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space 4 space straight i with hat on top space plus space 6 space straight j with hat on top space plus space straight lambda space straight k with hat on top
   (i)       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 3 cell negative 1 end cell row 4 6 straight lambda end table close vertical bar space equals space straight i with hat on top space open vertical bar table row 3 cell negative 1 end cell row 6 straight lambda end table close vertical bar space minus space straight j with hat on top open vertical bar table row 2 cell negative 1 end cell row 4 straight lambda end table close vertical bar space plus space straight k with hat on top space open vertical bar table row 2 3 row 4 6 end table close vertical bar
                      equals left parenthesis 3 straight lambda plus 6 right parenthesis space straight i with hat on top space minus space left parenthesis 2 straight lambda plus 4 right parenthesis straight j with hat on top plus left parenthesis 12 minus 12 right parenthesis straight k with hat on top space equals space left parenthesis 3 straight lambda plus 6 right parenthesis straight i with hat on top space minus space left parenthesis 2 space straight lambda space plus space 4 right parenthesis straight j with hat on top
Since straight a with rightwards arrow on top cross times straight b with rightwards arrow on top space equals space 0 with rightwards arrow on top
therefore space space space left parenthesis 3 straight lambda plus 6 right parenthesis straight i with hat on top space minus space left parenthesis 2 straight lambda plus 4 right parenthesis straight j with hat on top space equals space 0 with rightwards arrow on top therefore space space space 3 space straight lambda space plus space 6 space equals space 0 comma space space space 2 space straight lambda space plus space 4 space equals space 0 space space space rightwards double arrow space space space straight lambda space equals space minus 2
(ii) Since straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are perpendiuclar
therefore space space space straight a with rightwards arrow on top space straight b with rightwards arrow on top space equals space 0 therefore space space space space left parenthesis 2 right parenthesis thin space left parenthesis 4 right parenthesis plus left parenthesis 3 right parenthesis thin space left parenthesis 6 right parenthesis space plus space left parenthesis negative 1 right parenthesis space straight lambda space equals space 8 therefore space space space space 8 plus 18 minus straight lambda space equals space 0 space space space space space space space space space rightwards double arrow space space space space straight lambda space equals space 26
                  
                  

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