<p>Let <strong>a</strong>, <strong>b</strong>, <strong>c</strong> be unit vectors such that the Gram determinant
\[\begin{vmatrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} & \vec{a}\cdot\vec{c} \\ \vec{a}\cdot\vec{b} & \vec{b}\cdot\vec{b} & \vec{b}\cdot\vec{c} \\ \vec{a}\cdot\vec{c} & \vec{c}\cdot\vec{b} & \vec{c}\cdot\vec{c} \end{vmatrix} = [\vec{a}\;\vec{b}\;\vec{c}]^2 = 4.\]
Find \([\vec{a}\;\vec{b}\;\vec{c}]\).</p>
Step-by-Step Solution
Key Concept: The Gram determinant equals the square of the scalar triple product: G = [a b c]². Since a, b, c are unit vectors, the diagonal elements are all 1, and the determinant directly gives [a b c]². Taking the square root yields the scalar triple product.
Step 1: Recognize that a, b, c are unit vectors, so |a| = |b| = |c| = 1. Step 2: The Gram determinant for three vectors is defined as:
$G = \begin{vmatrix} \vec{a}\cdot\vec{a} & \vec{a}\cdot\vec{b} & \vec{a}\cdot\vec{c} \\ \vec{a}\cdot\vec{b} & \vec{b}\cdot\vec{b} & \vec{b}\cdot\vec{c} \\ \vec{a}\cdot\vec{c} & \vec{c}\cdot\vec{b} & \vec{c}\cdot\vec{c} \end{vmatrix}$ Step 3: Since a, b, c are unit vectors: $\vec{a}\cdot\vec{a} = \vec{b}\cdot\vec{b} = \vec{c}\cdot\vec{c} = 1$ Step 4: By the fundamental property of the Gram determinant:
$G = [\vec{a}\,\vec{b}\,\vec{c}]^2$
where $[\vec{a}\,\vec{b}\,\vec{c}]$ is the scalar triple product. Step 5: Given that $G = 4$:
$[\vec{a}\,\vec{b}\,\vec{c}]^2 = 4$
$[\vec{a}\,\vec{b}\,\vec{c}] = \pm 2$ Step 6: If the question expects a single numerical answer without sign specification, the magnitude is $|[\vec{a}\,\vec{b}\,\vec{c}]| = 2$. However, if directional orientation is considered, the answer can be either $+2$ or $-2$. The standard answer is ±2 , with the magnitude being 2 . Note: If the problem expects only the positive value or treats it as asking for magnitude: ∴ Answer: 2 (or ± 2 for the signed triple product)
Correct Answer: 4