α=i+2j+3k,β=2i+j−k
∗ indicates the cross product
α x β = (i+2j+3k) * (2i+j−k)
=2(i∗i)+(i∗j)−(i∗k)+4(j∗i)+2(j∗j)−2(j∗k)+6(k∗i)+3(k∗j)−3(k∗k)
since i∗i=j∗j=k∗k=0,i∗j=k,j∗k=i,k∗i=j,j∗i=−k,k∗j=−i,
i∗k=−j
so α x β =2(0)+k+j−4k+2(0)−2i+6j−3i−3(0)
=−5i+7j−3k
So option b is correct.
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