$\vec{a} = \hat{i} + \hat{j} + \lambda\hat{k}$ and $\vec{b} = \hat{i} - \hat{j}$, with $\lambda > 0$. The parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ has area $2\sqrt{3}$ square units. What is $\lambda$?

Explanation

1. 01

   $\vec{a} \times \vec{b} = \hat{i}(0 + \lambda) - \hat{j}(0 - \lambda) + \hat{k}(-1 - 1) = \lambda\hat{i} + \lambda\hat{j} - 2\hat{k}$.

2. 02

   Area $= \sqrt{2\lambda^{2} + 4} = 2\sqrt{3}$, so $2\lambda^{2} + 4 = 12$, $\lambda^{2} = 4$ and $\lambda = 2$.