Lebesgue measure of $A=\{\sum_i^n t_iv_i, 0\le t_i\le 1\}$

Question

Given are $v_1,...,v_n\in\mathbb R^n$, and the set $A=\{\sum_{i=1}^n t_iv_i, 0\le t_i\le 1\}$. How to calculate the Lebesgue measure $\lambda^n(A)$?

Can I solve it with applying $\lambda^n((a,b])=\Pi_i^n (a_i-b_i)$ $(*)$ ?

For example,$$v_1=(1,2,3)^T,v_2=(2,2,4)^T,v_3=(4,6,1)^T$$

how to apply $(*)$ ?

Answer 1

The solid $A$ is the parallelpiped spanned by the vectors $v_1,\ldots,v_n$. Its Lebesgue measure is the absolute value of the determinant of the matrix whose columns are the $v_i$: $$\lambda^n(A) = \bigg| \det \begin{bmatrix} v_1 | \cdots | v_n \end{bmatrix} \bigg|.$$