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Vector Algebra

Question
CBSEENMA12034138
Wired Faculty App

If with reference to the right handed system of mutually perpendicular unit vectors straight i with hat on top comma space space straight j with hat on top space and space straight k with hat on top comma space space straight alpha with rightwards arrow on top space equals space 3 straight i with hat on top minus straight j with hat on top comma space space straight beta with rightwards arrow on top space equals 2 straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top comma space then express straight beta with rightwards arrow on top in the form straight beta with rightwards arrow on top space equals space stack straight beta subscript 1 with rightwards arrow on top space plus space straight beta with rightwards arrow on top subscript 2 comma space space where space stack straight beta subscript 1 with rightwards arrow on top space is space parallel space to space straight alpha with rightwards arrow on top space and space stack straight beta subscript 2 with rightwards arrow on top space is perpendicular to straight alpha with rightwards arrow on top.

Solution

Here
      straight alpha with rightwards arrow on top space equals space 3 space straight i with hat on top minus straight j with hat on top comma space straight beta with rightwards arrow on top space equals 2 space straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top
Since stack straight beta subscript 1 with rightwards arrow on top is parallel to straight alpha with rightwards arrow on top
therefore space space stack straight beta subscript 1 with rightwards arrow on top space equals space straight lambda straight alpha with rightwards arrow on top space where space straight lambda space is space straight a space scalar. therefore space space stack straight beta subscript 1 with rightwards arrow on top space equals space 3 space straight lambda space straight i with hat on top space minus space straight lambda space straight j 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 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
Now,       straight beta with rightwards arrow on top space equals space straight beta with rightwards arrow on top subscript 1 space plus space stack straight beta subscript 2 with rightwards arrow on top space space space rightwards double arrow space space space space stack straight beta subscript 2 with rightwards arrow on top space equals space straight beta with rightwards arrow on top space minus stack straight beta subscript 1 with rightwards arrow on top
 therefore space space space stack straight beta subscript 2 with rightwards arrow on top space equals space left parenthesis 2 straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top right parenthesis space minus space left parenthesis 3 space straight lambda space straight i with hat on top space minus space straight lambda space straight j with hat on top right parenthesis or space space space space stack straight beta subscript 2 with rightwards arrow on top space equals space left parenthesis 2 minus 3 space straight lambda right parenthesis space straight i with hat on top space plus space left parenthesis 1 plus straight lambda right parenthesis straight j with hat on top space minus space 3 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 space space space... left parenthesis 2 right parenthesis
Since straight beta with rightwards arrow on top subscript 2 is perpendicular to straight alpha with rightwards arrow on top.
space therefore space space space straight alpha with rightwards arrow on top. space stack straight beta subscript 2 with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space 3 space left parenthesis 2 space minus space 3 space straight lambda right parenthesis space minus space left parenthesis negative 1 right parenthesis space left parenthesis 1 plus straight lambda right parenthesis plus left parenthesis 0 right parenthesis left parenthesis negative 3 right parenthesis space equals space 0 therefore space space space 6 minus 9 straight lambda minus 1 minus straight lambda equals space space 0 space rightwards double arrow space space space space 10 space straight lambda space equals space 5 space space space rightwards double arrow space space straight lambda space equals space 1 half therefore space space space from space left parenthesis 1 right parenthesis comma space space straight beta with rightwards arrow on top subscript 1 space equals space 3 over 2 straight i with hat on top space minus space 1 half straight j with hat on top
and from (2),  stack straight beta subscript 2 with rightwards arrow on top space equals space open parentheses 2 minus 3 over 2 close parentheses straight i with hat on top space plus space open parentheses 1 plus 1 half close parentheses space straight j with hat on top space minus space 3 space straight k with hat on top
or               stack straight beta subscript 2 with rightwards arrow on top space equals space 1 half straight i with hat on top space plus 3 over 2 straight j with hat on top space minus space 3 space straight k with hat on top
therefore space space straight beta with rightwards arrow on top space equals space open parentheses 3 over 2 straight i with hat on top space minus 1 half straight j with hat on top close parentheses space plus space open parentheses 3 over 2 straight i with hat on top space plus space 3 over 2 straight j with hat on top space minus space 3 space straight k with hat on top close parentheses 

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