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Q.

A→  =  2i^+j^,  B = 3j^−k^   and  C→  =  6i^−2k^.  Value of  A→−2B→+3C→ would be

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a

20i^ +5j^ +4k^

b

20i^ −5j^ −4k^

c

4i^ +5j^ +20k^

d

5i^ +4j^ +10k^

answer is B.

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Detailed Solution

A→−2B→ +3C→  = (2i^+j^)−2(3j^−k^)+3(6i^−2k^)= 2i^+j^−6j^+2k^+18i^−6k^= 20i^−5j^−4k^
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