Given v=(3,-2i,1+i) so (v bar )can be written as (3,2i,1-i).
u.(v bar )=(i, 2i,7).(3,2i,1-i)=(3i ,4 i^2 ,7-7i )= (3i ,-4 ,7-7i)
Similarly , w.u = (2-i ,2i ,2+7i).(i, 2i,7)=(2i-i^2 ,2i^2,14+49i )=(2i+1,-2 ,14+49i)
bar of w.u is given as (1-2i ,-2 ,14-49i )
Now , subtract (1-2i ,-2,14-49i ) from (3i,-4,7-7i) that gives us , (-1+5i ,-2,-7+42i )
Take the bar of (-1+5i ,-2 ,-7+42i) we will get (-1-5i ,-2,-7-42i ) as answer
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