Decompose vec d=(3vec a+vec b-2vec c)×(vec a-vec b+5vec c)…
Decompose $\vec d=(3\vec a+\vec b-2\vec c)\times(\vec a-\vec b+5\vec c)$ along the mutually perpendicular unit vectors $\vec a,\vec b,\vec c$, which form a right-handed triad.
Explanation of Method
Decomposing a vector product in an orthonormal right-handed basis.
Step 1: Determinant form
Since $\vec a\times\vec b=\vec c$, $\vec b\times\vec c=\vec a$, $\vec c\times\vec a=\vec b$, the vectors $\vec a,\vec b,\vec c$ play the role of $\vec i,\vec j,\vec k$: \[\vec d=\begin{vmatrix}\vec a&\vec b&\vec c\\3&1&-2\\1&-1&5\end{vmatrix}=\vec a(5-2)-\vec b(15+2)+\vec c(-3-1).\]
\[\vec d=3\vec a-17\vec b-4\vec c\]
Step 2: Check by term-by-term expansion
| Term | Value | Term | Value |
|---|---|---|---|
| $3\vec a\times(-\vec b)$ | $-3\vec c$ | $\vec b\times5\vec c$ | $5\vec a$ |
| $3\vec a\times5\vec c$ | $-15\vec b$ | $-2\vec c\times\vec a$ | $-2\vec b$ |
| $\vec b\times\vec a$ | $-\vec c$ | $-2\vec c\times(-\vec b)$ | $-2\vec a$ |
The remaining three terms are of the form $\vec u\times\vec u=\vec 0$. Collecting: $\vec a:\ 5-2=3$, $\vec b:\ -15-2=-17$, $\vec c:\ -3-1=-4$ $\checkmark$
Result
\[\boxed{\,\vec d=3\vec a-17\vec b-4\vec c\,}\]