Let a→=i^+j^+2k^,  b¯=b1i^+b2j^+2k^ andc→=5i→+j^+2k  be three vectors such that the projection of b→ on a→ is |a→|. If a→+b→ is perpendicularto c→, then |b→|  is equal to 

Let a=i^+j^+2k^,  b¯=b1i^+b2j^+2k^ and

c=5i+j^+2k  be three vectors such that the 

projection of b on a is |a|. If a+b is perpendicular

to c, then |b|  is equal to 

  1. A

    32

  2. B

    22

  3. C

    4

  4. D

    6

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    Solution:

    Given, projection of b¯ on a=a·b|a|=|a|

      b1+b2+24=4b1+b2=2

    Also, (a+b)c(a+b)·c=0

    1+b15+1+b21+22(2)=0

      5b1+b2=-10

    Solving  (i) and  (ii), we get b1=-3 and b2=5

    Now, |b|=b12+b22+2=6

     

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