Find the volume of the parallelepiped formed by the vectors $\mathbf{a}=(3,-2,5)$, $\mathbf{b}=(1,4,-1)$, and $\mathbf{c}=(-2,3,0)$.

Step-by-step solution

1. Set up the scalar triple product as a determinant:\n$\left|\begin{array}{ccccccc}3& -2& 5\text{1}& 4& -1\text{-}2& 3& 0\end{array}\right|$.
2. Expand along the first row: $3\left|\begin{array}{ccc}4& -1\text{3}& 0\end{array}\right|-(-2)\left|\begin{array}{ccc}1& -1\text{-}2& 0\end{array}\right|+5\left|\begin{array}{ccc}1& 4\text{-}2& 3\end{array}\right|$.
3. Compute each $2\times 2$ determinant: $4\cdot 0-(-1)\cdot 3=3$, $1\cdot 0-(-1)(-2)=-2$, $1\cdot 3-4(-2)=3+8=11$.
4. Substitute: $3\cdot 3+2\cdot (-2)+5\cdot 11=9-4+55=60$.
5. Take the absolute value to get the volume: $|60|=60$.
6. Thus the volume of the parallelepiped is $60$ cubic units.

Concept tested: Volume via scalar triple product

Published 15 September 2026 · How this solution was prepared and checked