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Vector Algebra

Question
CBSEENMA12034129

Find the scalar components of a unit vector which is perpendicular to the vectors straight i with hat on top plus 2 space straight j with hat on top space minus space straight k with hat on top space space and space space 3 straight i with hat on top minus straight j with hat on top plus 2 space straight k with hat on top.

Solution

Let straight r with rightwards arrow on top space equals straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top                      ...(1)
be the unit vector which is perpendicular to the vectors
          straight a with rightwards arrow on top space equals straight i with hat on top space plus space 2 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 3 straight i with hat on top space minus straight j with hat on top space plus 2 straight k with hat on top
therefore space space space space space straight x squared plus straight y squared plus straight z squared space equals 1                                                ...(2)
          straight x plus 2 straight y minus straight z space equals space 0                                                  ...(3)
           3 straight x minus straight y plus 2 straight z space equals space 0                                              ...(4)
Multiplying (3) by 1 and (4) by 2, we get,
x + 2y – z = 0
6x – 2y + 4z = 0
Adding,    7 straight x plus 3 straight z space equals space 0 comma space space space space therefore space space space straight z space equals space minus 7 over 3 straight x
Multiplying (3) by 2 and (4) by 1, we get,
2x + 4y – 2z = 0
3x – y + 2z = 0
Adding,       5 straight x plus 3 straight y space equals space 0 space space space space space space rightwards double arrow space space space straight y space equals space minus fraction numerator 5 straight x over denominator 3 end fraction
Putting     straight y space equals space minus fraction numerator 5 straight x over denominator 3 end fraction comma space space space straight z space equals space minus fraction numerator 7 straight x over denominator 3 end fraction space in space left parenthesis 1 right parenthesis comma space we space get comma
          straight x squared plus 25 over 9 straight x squared plus 49 over 9 straight x squared space equals space 1 space space space space space space rightwards double arrow space space space straight x space equals space space plus-or-minus fraction numerator 3 over denominator square root of 83 end fraction
therefore space space space straight x squared space equals space 9 over 83 space space space space space space space rightwards double arrow space space space straight x space equals space plus-or-minus fraction numerator 3 over denominator square root of 83 end fraction
therefore space space space straight y space equals plus-or-minus fraction numerator 5 over denominator square root of 83 end fraction comma space space straight z space equals space plus-or-minus fraction numerator 7 over denominator square root of 83 end fraction
therefore space space space required space scalar space components space are
space space space space space space space space space space space space space fraction numerator 3 over denominator square root of 83 end fraction comma space minus fraction numerator 5 over denominator square root of 83 end fraction comma space minus fraction numerator 7 over denominator square root of 83 end fraction space space space or space space space minus fraction numerator 3 over denominator square root of 83 end fraction comma space fraction numerator 5 over denominator square root of 83 end fraction comma space fraction numerator 7 over denominator square root of 83 end fraction