Question:
<p>If the vectors \(\vec{a}+2\vec{b}+3\vec{c}\), \(\vec{b}+2\vec{c}+3\vec{a}\), \(\vec{c}+2\vec{a}+3\vec{b}\) are coplanar, then</p>
\(\vec{a},\vec{b},\vec{c}\) are always coplanar
\(\vec{a},\vec{b},\vec{c}\) are never coplanar
\(\vec{a},\vec{b},\vec{c}\) are coplanar only if \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\)
\(\vec{a},\vec{b},\vec{c}\) may or may not be coplanar
Step-by-Step Solution
Key Concept: The three combined vectors can be expressed as M \cdot [a,b,c]ᵀ where M is a 3 \times 3 matrix with rows (1,2,3),(3,1,2),(2,3,1). det(M)=18\neq0, so the combined vectors are coplanar iff the originals are.
Write the three vectors as a matrix product:
$\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix} = \begin{pmatrix}1&2&3\\3&1&2\\2&3&1\end{pmatrix}\begin{pmatrix}\vec{a}\\\vec{b}\\\vec{c}\end{pmatrix}$
The matrix $M=\begin{pmatrix}1&2&3\\3&1&2\\2&3&1\end{pmatrix}$ has $\det(M)=1(1-6)-2(3-4)+3(9-2)=-5+2+21=18\neq0$.
Since $\det(M)\neq0$, the transformation is invertible. The combined vectors are coplanar iff the originals are. Answer: (A)
Correct Answer: A