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Vector Algebra

Question
CBSEENMA12034165
Wired Faculty App

Find straight lambda space and space straight mu if  left parenthesis 2 space straight i with hat on top space plus space 6 space straight j with hat on top space plus space 27 space straight k with hat on top right parenthesis space cross times left parenthesis straight i with hat on top space cross times space straight lambda space straight j with hat on top space plus space straight mu space straight k with hat on top right parenthesis space equals space 0 with rightwards arrow on top.

Solution

Here  left parenthesis 2 space straight i with hat on top space plus space 6 space straight j with hat on top space plus space 27 space straight k with hat on top right parenthesis space cross times space left parenthesis straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space straight mu space straight k with hat on top right parenthesis space equals space 0 with rightwards arrow on top
therefore space space space 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 6 27 row 1 straight lambda straight mu end table close vertical bar space equals space 0 with rightwards arrow on top therefore space space space space space straight i with hat on top space open vertical bar table row 6 27 row straight lambda straight mu end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 2 27 row 1 straight mu end table close vertical bar space plus space straight k with hat on top space open vertical bar table row 2 6 row 1 straight lambda end table close vertical bar space equals 0 with rightwards arrow on top therefore space space space space space space left parenthesis 6 space straight mu space minus space 27 space straight lambda right parenthesis straight i with hat on top space minus space left parenthesis 2 straight mu minus 27 right parenthesis space straight j with hat on top space plus left parenthesis 2 straight lambda minus 6 right parenthesis space straight k with hat on top space equals space 0 with rightwards arrow on top rightwards double arrow space space space space space space space space space 6 space straight mu space minus 27 space straight lambda space equals space 0 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 space space space space space space... space space left parenthesis 1 right parenthesis space space space space space space space space minus left parenthesis 2 straight mu minus 27 right parenthesis space equals space 0 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 space space space space space space space space space space... left parenthesis 2 right parenthesis and space space 2 space straight lambda minus 6 space equals space 0 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 space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis From space left parenthesis 3 right parenthesis comma space 2 space straight lambda space equals space 6 comma space space space space space space space space space space space rightwards double arrow space space straight lambda space equals space 3 From space left parenthesis 2 right parenthesis comma space 2 space straight mu space equals space 27 comma space space space space space space space space space rightwards double arrow space straight mu space equals space 27 over 2 Putting space space space space space straight lambda space equals space 3 comma space space space straight mu space equals space 27 over 2 space in space left parenthesis 1 right parenthesis comma space we space get comma space space space space space space 6 open parentheses 27 over 2 close parentheses space minus space 27 space cross times 3 space equals space 0 or space space space 81 minus 81 space equals space 0 or space space space space space space space space space space 0 space equals space 0 therefore space space space space space space space space space space space straight lambda space equals space 3 comma space space space straight mu space equals space 27 over 2 space satisfy space all space the space equations therefore space space space space space space space space space space space straight lambda space equals space 3 comma space space space straight mu space equals space 27 over 2

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