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Q.

lf a=i+j+k, b =-i+2j+k,  c=i+2j-k then  a unit vector perpendicular to a+ b and b+c is

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a

i+2j

b

i

c

j

d

j+k

answer is B.

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Detailed Solution

a+b=3j+2k     b+c=4j (a+b)×(b+c)=ijk032040=i(-8)-j(0)+k(0-0) =-8i |(a+b)×(b+c)|=8

Unit vector =±-8i8=±i

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