The vector product \((2\hat{\imath}+\hat{\jmath}-\hat{k})\times(\hat{\imath}-\hat{\jmath}+2\hat{k})\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\hat{\imath}-5\hat{\jmath}-3\hat{k}\)
Writing the vector product as a determinant:
\( \displaystyle \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & 2 \end{vmatrix} \)
\( = \hat{\imath}\,[(1)(2)-(-1)(-1)]-\hat{\jmath}\,[(2)(2)-(-1)(1)]+\hat{k}\,[(2)(-1)-(1)(1)]\)
\( = \hat{\imath}\,(2-1)-\hat{\jmath}\,(4+1)+\hat{k}\,(-2-1)\)
\( = \hat{\imath}-5\hat{\jmath}-3\hat{k}\)
Why the other options are incorrect
Option (2) \(-\hat{\imath}+5\hat{\jmath}+3\hat{k}\) — Incorrect. This is the product in the reverse order, \((\hat{\imath}-\hat{\jmath}+2\hat{k})\times(2\hat{\imath}+\hat{\jmath}-\hat{k})\).
Option (3) \(\hat{\imath}+5\hat{\jmath}-3\hat{k}\) — Incorrect. The minus sign in front of the \(\hat{\jmath}\) term is missed.
Option (4) \(3\hat{\imath}-5\hat{\jmath}-\hat{k}\) — Incorrect. The \(\hat{\imath}\) component is \(2-1 = 1\), not 3.