The value of \(\hat{\imath}\times(\hat{\jmath}+\hat{k})+\hat{\jmath}\times(\hat{k}+\hat{\imath})+\hat{k}\times(\hat{\imath}+\hat{\jmath})\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\vec{0}\)
In the cyclic order \(\hat{\imath}\to\hat{\jmath}\to\hat{k}\to\hat{\imath}\), each cross product gives the next vector, and reversing the order changes the sign (\(\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}\)):
\(\hat{\imath}\times\hat{\jmath} = \hat{k}, \quad \hat{\jmath}\times\hat{k} = \hat{\imath}, \quad \hat{k}\times\hat{\imath} = \hat{\jmath}\)
\(\hat{\imath}\times\hat{k} = -\hat{\jmath}, \quad \hat{\jmath}\times\hat{\imath} = -\hat{k}, \quad \hat{k}\times\hat{\jmath} = -\hat{\imath}\)
Using the distributive law on each term:
\(\hat{\imath}\times(\hat{\jmath}+\hat{k}) = \hat{k}-\hat{\jmath}\)
\(\hat{\jmath}\times(\hat{k}+\hat{\imath}) = \hat{\imath}-\hat{k}\)
\(\hat{k}\times(\hat{\imath}+\hat{\jmath}) = \hat{\jmath}-\hat{\imath}\)
Adding, every unit vector appears once with + and once with −:
\((\hat{k}-\hat{\jmath})+(\hat{\imath}-\hat{k})+(\hat{\jmath}-\hat{\imath}) = \vec{0}\)