The area of the parallelogram whose adjacent sides are \(\hat{\imath}+\hat{\jmath}\) and \(\hat{\jmath}+\lambda\hat{k}\) is 3 square units. The positive value of \(\lambda\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The area of the parallelogram is the magnitude of the cross product of its adjacent sides.
\( \displaystyle (\hat{\imath}+\hat{\jmath})\times(\hat{\jmath}+\lambda\hat{k}) = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & \lambda \end{vmatrix} \)
\( = \hat{\imath}(\lambda-0)-\hat{\jmath}(\lambda-0)+\hat{k}(1-0)\)
\( = \lambda\hat{\imath}-\lambda\hat{\jmath}+\hat{k}\)
Its magnitude must be 3:
\(\sqrt{2\lambda^2+1} = 3\)
\(2\lambda^2+1 = 9\)
\(\lambda^2 = 4\)
The positive value is \(\lambda = 2\).