Vector Algebra
Scalar Triple Product and Gram Determinant
Grade 12
Question:
<p>If \([\vec{a}+2\vec{b}+3\vec{c}\quad \vec{b}+2\vec{c}+3\vec{a}\quad \vec{c}+2\vec{a}+3\vec{b}] = 36\), where \(\vec{a},\vec{b}\) and \(\vec{c}\) are three vectors, then the value of \(\begin{vmatrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} & \vec{a}\cdot\vec{c} \\ \vec{a}\cdot\vec{b} & \vec{b}\cdot\vec{b} & \vec{b}\cdot\vec{c} \\ \vec{a}\cdot\vec{c} & \vec{c}\cdot\vec{b} & \vec{c}\cdot\vec{c} \end{vmatrix}\) is ________.</p>
Step-by-Step Solution
Key Concept: The scalar triple product [u v w] can be expressed as det(u v w), and if u, v, w are linear combinations of a, b, c, then [u v w] = det(coefficient matrix) × [a b c]. The Gram matrix determinant equals [a b c]².
\textbf{Step 1:} Express the given vectors as a linear combination. Let $\vec{u}_1 = \vec{a} + 2\vec{b} + 3\vec{c}$, $\vec{u}_2 = 3\vec{a} + \vec{b} + 2\vec{c}$, $\vec{u}_3 = 2\vec{a} + 3\vec{b} + \vec{c}$. The coefficient matrix is:
$$M = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix}$$
\textbf{Step 2:} Calculate $\det(M)$ using cofactor expansion:
$$\det(M) = 1(1 - 6) - 2(3 - 4) + 3(9 - 2) = 1(-5) - 2(-1) + 3(7) = -5 + 2 + 21 = 18$$
\textbf{Step 3:} Use the property $[\vec{u}_1 \; \vec{u}_2 \; \vec{u}_3] = \det(M) \times [\vec{a} \; \vec{b} \; \vec{c}]$, so:
$$36 = 18 \times [\vec{a} \; \vec{b} \; \vec{c}]$$
Therefore: $[\vec{a} \; \vec{b} \; \vec{c}] = 2$
\textbf{Step 4:} The Gram determinant (given matrix) equals $[\vec{a} \; \vec{b} \; \vec{c}]^2$:
$$\det(\text{Gram matrix}) = [\vec{a} \; \vec{b} \; \vec{c}]^2 = 2^2 = 4$$
$$\therefore \text{Answer: } 4$$
Correct Answer: 4