IF (10 x 8) + (r x 10) +[tex] {8}^{2} [/tex]-[tex] {r}^{2} [/tex]=100, then r =​

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1. Answer:

   So (r-c)×b =0

   Hence we can conclude that (r-c) is parallel vector to b,so

   r-c=Qb, where Q is a scaler.

   r=Qb +c……….(i)

   And we are provided with another condition that r.a=0

   So putting value of r from (i) we get,

   (Qb+c).a=0

   And proceed further.

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